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Chapter 9: Problem 38

Two identical blocks, each of mass \(M\), are connected by a light string over a frictionless pulley of radius \(R\) and rotational inertia \(I\) (Fig. 9-55). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and the sliding block. When this system is released, it is found that the pulley turns through an angle \(\theta\) in time \(t\) and the acceleration of the blocks is constant. (a) What is the angular acceleration of the pulley? ( \(b\) ) What is the acceleration of the two blocks? ( \(c\) ) What are the tensions in the upper and lower sections of the string? All answers are to be expressed in terms of \(M, I, R, \theta, g\), and \(t\).

Short Answer

Expert verified
The angular acceleration is \(2\theta / t^2\), the acceleration of the blocks is \(2R\theta / t^2\), and the tensions in the upper and lower sections of the string are \(Mg - 2MR\theta / t^2\) and \(Mg + 2MR\theta / t^2\) respectively.

Step by step solution

01

Find the angular acceleration of the pulley

The angular acceleration \(\alpha\) of the pulley is equal to the angle \(\theta\) divided by half the time squared. This comes from the equation \(\theta = \frac{1}{2} \alpha t^2\). Hence, \(\alpha = 2\theta / t^2\).
02

Find the acceleration of the two blocks

The acceleration \(a\) of the blocks is related to the angular acceleration of the pulley through the radius of the pulley. This is because the string does not slip on the pulley. Hence, acceleration \(a = \alpha R\). Substituting \(\alpha\) from Step 1, \(a = 2R\theta / t^2\).
03

Find the tensions in the upper and lower sections of the string

The tension \(T_1\) in the upper section of the string equals the force experienced by the falling block, which is its weight minus its force due to acceleration. Therefore, \(T_1 = Mg - Ma\). Substituting \(a\) from Step 2, \(T_1 = Mg - 2MR\theta / t^2\).The tension \(T_2\) in the lower section of the string is the sum of the weight of the block on the inclined plane and the force due to acceleration in the direction of motion. Therefore, \(T_2 = Mg + Ma\). Substituting \(a\) from Step 2, \(T_2 = Mg + 2MR\theta / t^2\).

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