Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion. Torque and angular momentum

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Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion Torque and angular momentum In Figure, in order to turn a rod about a fixed hinge at one end, a force F is applied at

Note that the perpendicular distance r has to be measured from the rotation axis to the line of force F. Consider a rotating body about a fixed axis as in Figure.

1 Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion Torque and angular momentum In Figure, in order to turn a rod about a fixed hinge at one end, a force F is applied at a point with position vector r relative to the hinge (pivot). Torque τ is defined by τ = r F, τ = rf sin θ, where θ is the angle between vector r and vector F. If θ = 90, torque τ = rf. Note that the perpendicular distance r has to be measured from the rotation axis to the line of force F. Consider a rotating body about a fixed axis as in Figure. Let I be moment of inertia and ω be the angular speed of the rotation about an axis. The angular momentum L is defined by L = Iω. Figure : Applying torque on an object Figure : Angular momentum of rigid body Note the direction of L coincides with the direction of ω. The angular momentum of the body changes upon at action of torque. Torque is equal to the rate of change of the angular momentum: τ = dl dt = d Iω dt = I dω dt = Iα, where α = dω dt is the angular acceleration. Consider Figure 3; a particle (point mass) travelling with velocity v at position vector r from O where the rotation axis passes through. The angular momentum of the particle is given by Figure 3: Angular momentum of a particle L = r p L = r mv, where p = mv is the linear momentum of the particle. If the velocity v is perpendicular to r, the formula for the angular momentum of a particle becomes. L = mvr.

2 Example A solid sphere of mass m = 0 kg and radius R = 0.35 m is rotating about axis through its center by the action three forces in the figure. a) Calculate the moment of inertia of this sphere. b) Calculate net torque and angular acceleration. Example A Bicycle brakes are being checked at a repair shop. The brakes deliver an 8-N tangential friction force to the.0-m diameter wheel rim If the wheel is rotating at 0 rad s - and its moment of inertia is.0 kg m, how long does it take for the wheel to come to rest? Example 3 A string is wrapped around a pulley, a disc of mass m and radius r, which can rotate about a fixed horizontal axis. A mass m hangs from the free end of the string. Show that the acceleration of the mass when it is let go is a = g/3.

3 3 Conservation of angular momentum Upon action of zero net torque, the total angular momentum L = Iω must be conserved: L before = L after. This is illustrated in Figure 3. In Figure 3(a), the ice skater has her arms extended, increasing the momentum of inertia of her body. By pulling in her arms in as in Figure 3(b), the moment of inertia decreases. Therefore, the angular speed of rotation increases. Example 4 A thinner disc of moment of inertia 0.55 kg m is slowly placed on a thicker rotating disc with moment of inertia of.05 kg m and angular speed of ω 0 =.0 rad s -. Calculate the angular speed ω f of the combined discs. Figure 4: Conservation of angular momentum Example 5 A merry-go-round has moment of inertia I = 00 kg m about its rotation axis. A kid running at speed v = ms - tangential to the rim jumps on the merry-go-round at distance r = m from the center. The kid is modeled as a particle of mass m = 40 kg. Determine the angular speed of the merry-go-round when the kid is on it.

4 4 Rotational kinetic energy Consider rotation of particles about a fixed axis in Figure 5. The total kinetic energy of the particles is E k = m i iv i. Since v i = ωr i, the above expression becomes E k = i m i ω r i = i m i r i ω = Iω. This equation can also be applied to a rigid body rotating about a fixed axis. The rotational kinetic energy of a rigid body is given by m 3 v 3 r 3 O r r v m m v Figure 5: Collection of particles rotating about fixed axis at O E rot = Iω. If there is not work done, the total energy of the system must be conserved. Example 6 A solid sphere of mass m = 5.0 kg and radius R = 0. m is rotating with frequency f = 0 Hz about a fixed axis passing through the center of the sphere. Calculate the rotational kinetic energy of the sphere. Example 7 A uniform rod of mass M and length L is pivoted at one end. The rod is released when it is horizontal. a) Determine the angular speed of the rod when the rod is vertical. b) Find the speed of the center of mass when the rod is vertical.

5 5 Rolling without slipping So far, the rotation axis is fixe (not moving). If the rotation axis is allowed to move in one dimension, this results in translational motion of the object as well as the rotational motion. The combination of these leads to rolling motion as illustrated in Figure 6. In Figure 6(a), an object undergoes pure translation and its center of mass is moving at speed v CM. All points on the object are moving at this speed. Figure 6(b) shows rotational motion about the center of mass with angular speed ω. The tangential speed at point P and P is v = Rω. Figure 6(c) is the combination of 6(a) and 6(b). If we impose a condition that the speed at point P is zero, we conclude that v CM = Rω This condition is known as rolling without slipping. When rolling without slipping occurs, rotation of a sphere of radius R by one revolution causes the center of mass moves a distance s = πr. There are two cases when the condition of rolling without slipping is not satisfied: Figure 6: Rolling motion as a combination of translation and rotation If v CM > Rω, the motion is rolling and skidding. s > πr If v CM < Rω, the motion is rolling and slipping. s < πr In rolling motion, the total kinetic energy is composed of rotational kinetic energy and translational kinetic energy: E k = E rot + E trans = Iω + mv CM. Example 8 A solid sphere is rolling without slipping from height H down the slope. Show that the speed of the center of mass when the sphere is at the bottom of the slope is v = 0 7 gh.

By considering forces acting on the cylinder, show that the acceleration of the center of mass a

6 Example 9 A cylinder is rolling without slipping down an inclined plane making angle θ with the horizontal. By considering forces acting on the cylinder, show that the acceleration of the center of mass a = 3 g sin θ. 6 Example 0 A pool ball of radius r is at rest on point P. Find the height, in terms of r, at which the cue tip must hit the ball so that the ball rolls without slipping.

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013 PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be