Problem 9-3. (See figure on page 265 of text book) A force of 110 N is applied perpendicularly to the left edge of the rectangle. (a) Find the torque (magnitude and direction) produced by this force with respect to an axis perpendicular to the plane of the rectangle at corner A and (b) with respect to a similar axis at corner B.
(a) The lever arm is 0.76 m, so the magnitude of the torque is 110N × 0.76 m = 83.6 N-m. The torque tends to produce a counter-clockwise rotation, so it is positive.
(b) In this case, the lever arm is 1.4 m. The magnitude of the torque is 110N × 1.4 m = 154 N-m. The torque tends to produce a clockwise rotation, so it is negative.
Problem 9-12. Suppose you need to measure the mass of an object, but have only a 10.0-kg standard mass and a meter stick. Place a knife edge under the 50-cm mark of the meter stick. Then put the 10.0-kg standard mass at the 15-cm mark and place the object so as to balance the standard mass. If the system is balanced when the object is at the 72-cm mark, find the unknown mass.
Since the meter stick itself is balanced if the knife edge is at the 50-cm mark, you need only consider the torques exerted by the other masses. The standard mass exerts a counterclockwise torque of 10.0 kg × 9.8 m/s2 × 0.35 m. (0.35 m is the distance from the knife edge to the mass. The unknown mass exerts a clockwise torque of m × 9.8 m/s2 × 0.22 m. These must be equal in magnitude, so
m × 9.8 m/s2 × 0.22 m = 10.0 kg × 9.8 m/s2 × 0.35 m
m = 10.0 kg × 0.35 / 0.22
m = 16 kg.
Problem 9-31. The blades of a ceiling fan have a total moment of inertia of 0.16 kg-m2 and an angular acceleration of 7.0 rad/s2. What net torque is being applied to the blades?
Problem 9-46. A baseball is thrown such that the translational speed of its center of mass is 31 m/s, and its angular speed about the center of mass is 180 rad/s. Treat the baseball as if it were a uniform solid sphere of radius 3.7 cm. What fraction of the ball's total kinetic energy is rotational kinetic energy?
The translational kinetic energy is
The moment of inertia is I = (2/5)mR2, and the rotational kinetic energy is
The ratio of the rotational kinetic energy to the total kinetic energy is
Problem 9-58. A playground carousel is free to rotate about its center on frictionless bearings. The carousel has an angular speed of 3.14 rad/s, a moment of inertia of 125 kg-m2, and a radius of 1.50 m. A 40.0-kg person, standing still next to the carousel, jumps on to it very close to the outer edge. Find the resulting angular speed of the carousel and person.
The initial angular momentum of the carousel is
When the person has jumped on to the carousel, their moment of inertia is that of a mass m moving in a circle of radius r, mr2. The final momentum is
and this must have the same value as the initial angular momentum. So
