Physics 2 Dynamics of Rotational Motion Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Torque Torque is what causes angular acceleration (just like a force causes linear acceleration) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Torque Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Pivot Point

FA FC FB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Torque Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Pivot Point FA Force B will tend to rotate the bolt clockwise, which will tighten it. FC

FB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Torque Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Pivot Point FA Force B will tend to rotate the bolt clockwise, which will tighten it. Notice that force A will tend to rotate the bolt counter-clockwise, loosening it. What does force C do? FC

FB Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Torque Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Pivot Point FA Force B will tend to rotate the bolt clockwise, which will tighten it. Notice that force A will tend to rotate the bolt counter-clockwise, loosening it. What does force C do? FC

FB Force C doesnt cause any rotation at all there is no torque generated by force C. Why not? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Torque Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque to be applied to an object, there needs to be a force that acts at some distance away from a pivot point. For example, consider tightening a bolt with a wrench. Which of the 3 forces shown will tighten the bolt? Pivot Point FA Force B will tend to rotate the bolt clockwise, which will tighten it. Notice that force A will tend to rotate the bolt counter-clockwise, loosening it.

What does force C do? FC FB Force C doesnt cause any rotation at all there is no torque generated by force C. Why not? Force C points directly at the pivot point no torque is created in this case. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Torque Formula for torque: F r sin( ) This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB Torque Formula for torque: F r sin( ) Pivot Point This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). FA r Take a look at the diagram r and are shown for force A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Torque Formula for torque: F r sin( ) Pivot Point This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). FA r Take a look at the diagram r and are shown for force A. There are 2 ways to interpret the formula. If you group the Fsin() together, that represents the component of the force that is perpendicular to the radius. To get the most torque, the force should be applied perpendicular (can you see why from the formula?) Prepared by Vince Zaccone For Campus Learning

Assistance Services at UCSB Torque Lever Arm Formula for torque: F r sin( ) Pivot Point This is really just the magnitude of the torque. The angle in the formula is between the force and the radius (from the pivot point to where the force is applied). FA r Take a look at the diagram r and are shown for force A. There are 2 ways to interpret the formula. If you group the F sin() together, that represents the component of the force that is perpendicular to the radius. To get the most torque, the force should be applied perpendicular

(can you see why from the formula?) The other option is to group the r sin() together and call it the lever arm for the force. Think of this as the shortest distance from the pivot point to where the force is applied. This is the effective radius of the force. Again, to get maximum torque the angle should be 90. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the torque of each force shown with respect to the pivot point at the left end of the 2m long rod. F2=30N F1 is applied at the right end, and F2 is at the center. 120 50 F1=20N Prepared by Vince Zaccone For Campus Learning

Assistance Services at UCSB Example: Find the torque of each force shown with respect to the pivot point at the left end of the 2m long rod. F2=30N F1 is applied at the right end, and F2 is at the center. 120 50 We can simply use our definition of torque here. F1=20N F r sin( ) 1 (20N) (2m) sin(50 ) 30. 6N m 2 (30N) (1m) sin(120 ) 26. 0N m Notice the sign convention: Counter-clockwise torque is positive. Clockwise torque is negative. Prepared by Vince Zaccone For Campus Learning

Assistance Services at UCSB Torque We mentioned earlier that torques produce angular accelerations. We have a formula for this relationship: I This is really just Newtons 2nd law applied to rotational motion. The moment of inertia, I, takes the place of the mass, and we use angular acceleration instead of linear. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the angular acceleration of the 2m long, uniform rod (mass=5kg) when it is subject to the 2 forces shown. F2=30N F1 is applied at the right end, and F2 is at the center. 120 50

This is just like the last problem, so we can use the results here. We need to add up all the torques on the rod, then solve for . F1=20N Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the angular acceleration of the 2m long, uniform rod (mass=5kg) when it is subject to the 2 forces shown. F2=30N F1 is applied at the right end, and F2 is at the center. 120 50 This is just like the last problem, so we can use the results here. We need to add up all the torques on the rod, then solve for . I 1 2 ( 13 ML2 )( )

F1=20N Look up this formula for the moment of inertia of a rod, with the axis at the end. 30. 6N m 26. 0N m ( 13 ) (5kg) (2m)2 . 69 rad s2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Angular Momentum This is the rotational equivalent of linear momentum. It quantifies the momentum of a rotating object, or system of objects. If we simply translate the variables, we can get one possible equation: p m v This is the formula for linear momentum

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Angular Momentum This is the rotational equivalent of linear momentum. It quantifies the momentum of a rotating object, or system of objects. If we simply translate the variables, we can get one possible equation: p m v This is the formula for linear momentum L I Using I instead of m and instead of v, we get this formula for angular momentum Prepared by Vince Zaccone For Campus Learning

Assistance Services at UCSB Angular Momentum This is the rotational equivalent of linear momentum. It quantifies the momentum of a rotating object, or system of objects. If we simply translate the variables, we can get one possible equation: p m v This is the formula for linear momentum L I Using I instead of m and instead of v, we get this formula for angular momentum This formula will work for a rigid body rotating about an axis of symmetry. In this case the angular momentum will be directed along the rotation axis. Use a right-hand rule to determine which direction. The more general formula for a particle is below:

= Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Angular Momentum Consider the simple case of a small mass (m) tied to a string with radius r. If the mass is swung around in a circle it will have some angular velocity . Notice that it will also have linear velocity (tangential to the circle). The relationship we know for these is v=r. m We can use this idea to find a useful alternate r formula for angular momentum. The point mass will have a moment of inertia: I m r 2 Substituting into the standard formula: L I

v L (m r ) ( ) r Lpo int mass m v r 2 We can use this formula when we have a point mass with a given linear velocity at some distance from a pivot point or axis of rotation. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Angular Momentum Like linear momentum, angular momentum is conserved. We will use this concept in several types of problems. I f f I i i This is a formula for conservation of angular momentum. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Angular Momentum Like linear momentum, angular momentum is conserved. We will use this concept in several types of problems. I f f I i i This is a formula for conservation of angular momentum. Also, we should have some formula relating angular momentum to torque (just like we have a formula relating linear momentum to force): = This says that any time a torque is applied, there will be a corresponding change in angular momentum. We can usually keep track of the vector aspect of this with +/- signs. Typically the positive direction is defined to coincide with a counter-clockwise rotation. Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB Example: A 28 kg child is sitting on the edge of a 145 kg merry-go-round of radius 2.5 m while it is spinning at a rate of 3.7 rpm. If the child moves to the center, how fast will it be spinning? Assume the merry-go-round is a uniform cylinder. initial final r=2.5m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 28 kg child is sitting on the edge of a 145 kg merry-go-round of radius 2.5 m while it is spinning at a rate of 3.7 rpm. If the child moves to the center, how fast will it be spinning? Assume the merry-go-round is a uniform cylinder. We can use conservation of momentum for this one. Initially both the merry-go-round and the child contribute to the total angular momentum.

However, once the child is at the center, she no longer has angular momentum (r=0). initial final r=2.5m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 28 kg child is sitting on the edge of a 145 kg merry-go-round of radius 2.5 m while it is spinning at a rate of 3.7 rpm. If the child moves to the center, how fast will it be spinning? Assume the merry-go-round is a uniform cylinder. We can use conservation of momentum for this one. Initially both the merry-go-round and the child contribute to the total angular momentum. However, once the child is at the center, she no longer has angular momentum (r=0). L f Li initial We need the total angular momentum of the system,

Including both the child and the disk. I disk, f disk, f I disk,i disk,i I girl,i girl,i final r=2.5m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 28 kg child is sitting on the edge of a 145 kg merry-go-round of radius 2.5 m while it is spinning at a rate of 3.7 rpm. If the child moves to the center, how fast will it be spinning? Assume the merry-go-round is a uniform cylinder. We can use conservation of momentum for this one. Initially both the merry-go-round and the child contribute to the total angular momentum. However, once the child is at the center, she no longer has angular momentum (r=0). L f Li initial

We need the total angular momentum of the system, Including both the child and the disk. I disk, f disk, f I disk,i disk,i I girl,i girl,i final r=2.5m The initial angular velocity is given in rpm, and we can leave it in those units for this type of problem. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 28 kg child is sitting on the edge of a 145 kg merry-go-round of radius 2.5 m while it is spinning at a rate of 3.7 rpm. If the child moves to the center, how fast will it be spinning? Assume the merry-go-round is a uniform cylinder. We can use conservation of momentum for this one. Initially both the merry-go-round and the child contribute to the total angular momentum. However, once the child is at the center, she no longer has angular momentum (r=0).

L f Li initial We need the total angular momentum of the system, Including both the child and the disk. I disk, f disk, f I disk,i disk,i I girl,i girl,i final r=2.5m The initial angular velocity is given in rpm, and we can leave it in those units for this type of problem. 2 2 2 ( 12 mdiskrdisk ) disk, f ( 12 mdiskrdisk ) ( i ) (mgirlrgirl ) ( i ) ( 12 145kg (2. 5m)2 ) disk, f ( 12 145kg (2. 5m)2 ) (3. 7rpm) (28kg (2. 5m)2 ) (3. 7rpm)

disk, f 5. 1rpm Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A helicopter rotor blade can be considered a long thin rod, as shown in figure below. Each of the three rotor helicopter blades is L = 3.75 m long and has a mass of m = 160 kg. While starting up, the motor provides a constant torque for 2 minutes, and the blades speed up from rest to 85 rad/s. How much torque is produced by the motor? How fast is the tip of each blade moving? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A helicopter rotor blade can be considered a long thin rod, as shown in figure below. Each of the three rotor helicopter blades is L = 3.75 m long and has a mass of m = 160 kg. While starting up, the motor provides a constant torque for 2 minutes, and the blades speed up from rest to 85 rad/s.

How much torque is produced by the motor? How fast is the tip of each blade moving? We will use angular momentum for this one. Since we have a constant torque, we simply need to find the change in angular momentum and divide by time. = Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A helicopter rotor blade can be considered a long thin rod, as shown in figure below. Each of the three rotor helicopter blades is L = 3.75 m long and has a mass of m = 160 kg. While starting up, the motor provides a constant torque for 2 minutes, and the blades speed up from rest to 85 rad/s. How much torque is produced by the motor? How fast is the tip of each blade moving? We will use angular momentum for this one. Since we have a constant torque, we simply need to find the change in angular momentum and divide by time.

= Use the definition of angular momentum: L I Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A helicopter rotor blade can be considered a long thin rod, as shown in figure below. Each of the three rotor helicopter blades is L = 3.75 m long and has a mass of m = 160 kg. While starting up, the motor provides a constant torque for 2 minutes, and the blades speed up from rest to 85 rad/s. How much torque is produced by the motor? How fast is the tip of each blade moving? We will use angular momentum for this one. Since we have a constant torque, we simply need to find the change in angular momentum and divide by time.

= Use the definition of angular momentum: We need to find the moment of inertia for the blades. There are 3 of them, and each one is a long thin rod. L I I rod 13 ML2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A helicopter rotor blade can be considered a long thin rod, as shown in figure below. Each of the three rotor helicopter blades is L = 3.75 m long and has a mass of m = 160 kg. While starting up, the motor provides a constant torque for 2 minutes, and the blades speed up from rest to 85 rad/s. How much torque is produced by the motor? How fast is the tip of each blade moving?

We will use angular momentum for this one. Since we have a constant torque, we simply need to find the change in angular momentum and divide by time. = Use the definition of angular momentum: We need to find the moment of inertia for the blades. There are 3 of them, and each one is a long thin rod. L I I rod 13 ML2 L 3( 13 ML2 ) (160kg)( 3. 75m)2 (85 rad ) 191,250 s Since the blades started from rest, this is L.L. kg m2 s

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A helicopter rotor blade can be considered a long thin rod, as shown in figure below. Each of the three rotor helicopter blades is L = 3.75 m long and has a mass of m = 160 kg. While starting up, the motor provides a constant torque for 2 minutes, and the blades speed up from rest to 85 rad/s. How much torque is produced by the motor? How fast is the tip of each blade moving? We will use angular momentum for this one. Since we have a constant torque, we simply need to find the change in angular momentum and divide by time. = Use the definition of angular momentum: We need to find the moment of inertia for the blades. There are 3 of them, and each one is a long thin rod.

L I I rod 13 ML2 L 3( 13 ML2 ) (160kg)( 3. 75m)2 (85 rad ) 191,250 s Since the blades started from rest, this is L.L. 2 191,250 kg sm 1,594N m 120 s kg m2 s Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Example: A helicopter rotor blade can be considered a long thin rod, as shown in figure below. Each of the three rotor helicopter blades is L = 3.75 m long and has a mass of m = 160 kg. While starting up, the motor provides a constant torque for 2 minutes, and the blades speed up from rest to 85 rad/s. How much torque is produced by the motor? How fast is the tip of each blade moving? We will use angular momentum for this one. Since we have a constant torque, we simply need to find the change in angular momentum and divide by time. = Use the definition of angular momentum: We need to find the moment of inertia for the blades. There are 3 of them, and each one is a long thin rod. L I I rod 13 ML2

L 3( 13 ML2 ) (160kg)( 3. 75m)2 (85 rad ) 191,250 s Since the blades started from rest, this is L.L. 2 191,250 kg sm 1,594N m 120 s To find the speed of the blade tips, simply use v=r: v (3. 75m) (85 rad ) s kg m2 s v 319 ms Prepared by Vince Zaccone

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