Rotational Dynamics

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

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 where the force is applied: Torque F r axis

Magnitude of torque: Magnitude of torque

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… Decreases Increases 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… Decreases Increases Remains the same • A rotating bar r φ F and φ decreases

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… Decreases Increases Remains the same • A rotating bar r F φ F and φ decreases Lever arm decreases

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 Torque

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

Problem 10.2 16 N m 19 N m 28 N m 52 N m 64 N m

Problem 10.2 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!

No total torque = no angular acceleration Rotational Dynamics • 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 2nd law for translational motion • For rotations, both the moment of inertia and the torques depend on the axis (which must be same!)

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 F2 compare to F1? Assume that all the mass of the wheels is concentrated in the rims. F2 = ½F1 F2 = F1 F2 = 2F1 F2 = 4F1 None of the above Accelerating two wheels

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… less than mgR equal to mgR greater than mgR impossible to predict Example R I m

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 less than mgR equal to mgR greater than mgR impossible to predict Example R I m T

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 less than mgR- because T < mg! equal to mgR greater than mgR impossible to predict Example R I m T a mg

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? Example R I m Hanging block: α Pulley: T T Constraint: a mg

R I m α T T a mg

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

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 R2 (large), resulting in a very small angular acceleration Another example: Yo-yo ICM

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 R2 (large), resulting in a very small angular acceleration Acceleration: Another example: Yo-yo ICM

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

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

Rotational Dynamics