Consider the 12.0 kg motorcycle wheel shown in Figure 10.38. Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?
Step-by-step solution Step 1 of 3 Refer to the figure 10.38 provided in the textbook. First Calculate the torque and moment of inertia of wheel after that calculate angular acceleration. (a) Drive chain exerts a force on a radius, and then the torque generated is, Here, is the torque generated, is the force exerted, and is the radius. Substitute, and Suppose wheel as an annular ring, so the moment of inertia is calculated according to outer and inner radius of the wheel, From the moment of inertia of an annular ring is, Here is the moment of inertia of the ring is the mass of the wheel and is the inner radius and is the outer radius of the wheel. Substitute, and From the relation of torque moment of inertia and angular acceleration, calculate angular acceleration. Here is the angular acceleration of the wheel, Substitute, and in the above equation to get, Hence, angular acceleration of the wheel is .