Racquetball Striking a Wall Rotational Dynamics Stewart Hall

Racquetball Striking a Wall Rotational Dynamics Stewart Hall General Physics I Mechanics Physics 140 Mt. Etna

Torque • • • The “rotating effect” of a force is greater, if the force is greater The “rotating effect” of a force is greater, if the force is applied further from the axis of rotation The “rotating effect” of the force also depends on the direction of the force: it is maximized by a perpendicular force; no effect of a parallel force These three properties lead to the following definition of the torque: Vector r is directed from the axis to the point F where the force is applied: axis r

Magnitude of torque • Magnitude of torque:

A rotating bar • A mass is hanging from the end of a horizontal bar which pivots about an axis through it center, but is being held stationary. The bar is released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar… A. Decreases B. Increases C. Remains the same

A rotating bar • A mass is hanging from the end of a horizontal bar which pivots about an axis through it center, but is being held stationary. The bar is released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar… A. Decreases B. Increases C. Remains the same r φ F and φ decreases

A rotating bar • A mass is hanging from the end of a horizontal bar which pivots about an axis through it center, but is being held stationary. The bar is released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar… A. Decreases B. Increases C. Remains the same Lever arm decreases r F φ F and φ decreases

Torque • Depends on the axis

Torque • • Depends on the axis Depends on where the force is applied

Torque • • • Depends on the axis Depends on where the force is applied Specifically, gravitational force (weight) can be viewed as being applied at the center of mass

Torque • • Depends on the axis Depends on where the force is applied Specifically, gravitational force (weight) can be viewed as being applied at the center of mass What about the direction?

Torque • • • Depends on the axis Depends on where the force is applied Specifically, gravitational force (weight) can be viewed as being applied at the center of mass What about the direction? As usual for any vector product, direction of the torque is given by the right-hand rule:

Torque • • • Depends on the axis Depends on where the force is applied Specifically, gravitational force (weight) can be viewed as being applied at the center of mass What about the direction? As usual for any vector product, direction of the torque is given by the right-hand rule: In practice, we will be speaking about clockwise and counterclockwise torques

The Right Hand Rule: Calculating a Torque r rotates CCW into F F τ points out of page r r r Axis points out of page Axis points into page r F r rotates CW into F τ points into page

Problem 10. 2 A. B. C. D. E. 16 N m 19 N m 28 N m 52 N m 64 N m

Problem 10. 2 A. B. C. D. E. 16 N m 19 N m 28 N m 52 N m 64 N m |τ1| = 40 N m |τ2| = 12 N m The two torques are in the opposite directions!

Rotational Dynamics • No total torque = no angular acceleration • If the total torque exerted by all forces on a rigid body around some axis is not zero, then the angular acceleration of the body is given by where I is the moment of inertia around the same axis • Analogous to the Newton’s 2 nd law for translational motion • For rotations, both the moment of inertia and the torques depend on the axis (which must be same!)

Accelerating two wheels • A. B. C. D. E. Two wheels with fixed axles, each have the same mass M, but wheel 2 has twice the radius of wheel 1. Each is accelerated from rest with a force applied as shown. In order to impart identical angular accelerations to both wheels, how should F 2 compare to F 1? Assume that all the mass of the wheels is concentrated in the rims. F 2 = ½F 1 F 2 = 2 F 1 F 2 = 4 F 1 None of the above

Example • A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free to rotate around an axle passing through its center of mass. The rope does not slip on the disk. Because of gravity the block m falls and the pulley rotates. The magnitude of the torque on the pulley is… R A. less than mg. R I B. equal to mg. R C. greater than mg. R m D. impossible to predict

Example • A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free to rotate around an axle passing through its center of mass. The rope does not slip on the disk. Because of gravity the block m falls and the pulley rotates. The magnitude of the torque on the pulley is… = TR R A. less than mg. R T I B. equal to mg. R C. greater than mg. R m D. impossible to predict

Example • A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free to rotate around an axle passing through its center of mass. The rope does not slip on the disk. Because of gravity the block m falls and the pulley rotates. The magnitude of the torque on the pulley is… = TR R A. less than mg. R - because T < mg! I B. equal to mg. R T C. greater than mg. R m a D. impossible to predict mg

Example • A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free to rotate around an axle passing through its center of mass. The rope does not slip. What are the acceleration of the block and the tension in the string? Hanging block: R α Pulley: T I T a m mg Constraint:

R T I T a m mg α

Another example: Yo-yo • Work done by gravity goes not only to translational motion, but also to rotation: GPE = KEtrans + KErot and KErot >> KEtrans ICM

Another example: Yo-yo • • Work done by gravity goes not only to translational motion, but also to rotation: GPE = KEtrans + KErot and KErot >> KEtrans Alternatively, the torque is proportional to r (small), while the moment of inertia is proportional to R 2 (large), resulting in a very small angular acceleration ICM

Another example: Yo-yo • • • Work done by gravity goes not only to translational motion, but also to rotation: GPE = KEtrans + KErot and KErot >> KEtrans Alternatively, the torque is proportional to r (small), while the moment of inertia is proportional to R 2 (large), resulting in a very small angular acceleration Acceleration: ICM

Dynamics of a yoyo Translational motion: Rotational motion: T Solving for acceleration: r mg a M, I

Dynamics of a yoyo Approximate the yo-yo as a uniform disk of radius R: Then: T R r mg M, I