Assignment 9 1. A heavy cylindrical container is being rolled up an incline as shown, by applying a force parallel to the incline. The static friction coefficient is µ s. The cylinder has radius R, mass m and moment of inertia about its symmetry axis I. F d θ Draw the cylinder and the incline and make a free-body diagram showing all the forces on the cylinder and where they are applie Assume the motion up the incline is at constant speed, and that d > R. What direction is the static friction force? How do you know? Assuming µ s is large enough, or θ is small enough, to allow the cylinder to roll without slipping, how large must F be? Ans: F = R d mgsinθ. 2. In the previous problem: What is the best value for d, i.e., the value that lets F be as small as possible? Show that the minimum value of F is 1/2 the force one would need to push the cylinder up a frictionless incline of the same angle θ. Suppose the person applies this minimum force to roll the cylinder up the incline. What is the largest angle of incline that can be used, in terms of µ s? Ans: tanθ = 2µ s. 1
3. This situation is like one in Assignment 3, except that now the pulley s inertia must be taken into account. The incline is frictionless, the string is ideal, but the pulley has radius R and moment of inertia I about its frictionless axle. Call T 1 the tension in the string attached θ M I m to M, and T 2 the tension in the string attached to m. If m is large enough to fall with acceleration a, what is a? Ans: a = m Msinθ m + M + I /R 2 g Find the difference between the tensions, T 2 T 1. Ans: T 2 T 1 = (I /R 2 ) m Msinθ m + M + I /R 2 g. 4. In the situation of the previous problem, answer the same two questions: If the incline has friction, with kinetic coefficient µ k. Ans: a = m M(sinθ + µ k cosθ) m + M + I /R 2 g, T 2 T 1 = (I /R 2 ) m M(sinθ + µ k cosθ) m + M + I /R 2 g. If the incline is frictionless but the axle of the pulley has friction, which causes a torque τ f opposite to the angular velocity of the pulley. Ans: a = (m Msinθ)g τ f /R m + M + I /R 2, T 2 T 1 = (τ f /R) + (I /R 2 ) (m Msinθ)g τ f /R m + M + I /R 2. 2
5. You are to analyze how a car gets its energy from the torque the engine applies to the drive wheels. A torque supplies power according to the formula P = τω (derived from P = Fv for the power supplied by a force). Shown is the car, which is accelerating to the right. Its kinetic energy consists of three parts: the CM energy 1 2 Mv CM 2, the rotation energy of the front wheels 1 2 I 1 ω 2, and the rotation energy of the rear wheels 1 2 I 2 ω 2. We assume the wheels have the same radius R and are not slipping, so we have v CM = Rω. Let the front wheels be the drive wheels. A clockwise torque τ 0 is applied to them by the engine. This makes the point of contact of those wheels tend to slip backwards, so there is a static friction force f 1 to the right on them. What is the total torque on the front wheels? [It must be clockwise if the car is accelerating to the right.] The rear wheels must also have a clockwise angular acceleration so they don t slip. What direction is the static friction force f 2 from the road on those wheels? The friction forces do two things: they exert torques on their wheels and together they give the external force that accelerates the CM. Write the three equations expressing these things. [Be sure to include τ 0 in the torque on the front wheels.] Now calculate dk /dt, where K is the total kinetic energy, and show that dk /dt = τ 0 ω. This proves that the car s increase in energy comes from the engine s torque, even though that is internal to the car. 3
6. The refrigerator shown is being pushed by a horizontal force F applied along a line through its geometric center, which is also its CM. The static friction coefficient between the refrigerator and the floor is µ s. The dimensions of the refrigerator are F w h as shown, and its mass is m. If the refrigerator does not tip over, how large must F be to make it slide? Ans: F > µ s mg. Consider torques about the bottom right corner. For what value of F will the refrigerator be just on the verge of tipping over, if it doesn t slide? [The normal force from the floor must act effectively at some point on the bottom surface of the refrigerator.] Ans: F = (w/h)mg. Draw a line, starting at the CM, along the direction of the sum of F and mg. Show that if that line passes to the right of the bottom right corner, the refrigerator will tip over. What is the minimum value of µ s such that the refrigerator will slide rather than tip over. Ans: µ s < w/h. 7. Suppose the refrigerator in the previous problem is resting on the bed of a truck that is accelerating to the left as shown. The coefficient of static friction with the bed of the truck is µ s as before. Discuss the ways in which this situation is like the one in the previous problem. a w h Find the minimum value of a that will make the refrigerator slide if it doesn t tip over. Ans: a > µ s g. Find the value of a for which it is on the verge of tipping over, if it doesn t slide. Ans: a > (w/h)g. [The easiest approach is to use g eff.] 4
PHY 53 Summer 2010 Duke Marine Laboratory 8. The bowling ball shown starts down the alley with CM speed v 0, sliding but not rotating. The ball has mass m, radius R, and moment of inertia about its center 2 5 mr2. The coefficient of kinetic friction with the alley is µ k. What is the acceleration a (magnitude and direction) of the CM while the ball is sliding? Ans: a = µ k g, to left. What is the angular acceleration α (magnitude and direction) about the CM while the ball is sliding? [What direction is the torque?] Ans: α = 5 2 µ k (g/r), clockwise. Write the equations for the CM velocity v(t) while the ball is sliding. Ans: v(t) = v 0 µ k gt. Write the equation for the angular velocity ω(t) while the ball is sliding. [Take clockwise to be positive.] Ans: ω(t) = 5 2 µ k (g/r) t 9. The ball in the previous problem will stop sliding and roll instead when its linear and angular velocities obey the rolling condition. At what time will that happen? Ans: t = 2v 0 7µ k g. What is the CM speed when the ball rolls? Ans: v = 5 7 v o. How far did the ball slide before rolling? Ans: x = 12v 0 2 49µ k g. 1