Question

. Two identical blocks, each of mass M, are connected by a massless string over a pulley of radius R and moment of inertia I. The string does not slip on the pulley. It is not known whether there is friction between the table and the sliding block. The pulley's axis is, however, frictionless. When the system is released, it is found that the pulley turns through and angle θ in time t and the acceleration of the block 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? Express the answer in terms of M, I, R, θ, g, and t. 乙 R, I

Answer 1

The angular acceleration of a pulley is described under the equation,

$\theta = (0.5) \alpha t^2$

So things clearing,

$\alpha = \frac{ 2\theta}{t^2}$

b) For point B it is necessary to make a sum of Forces,

$\sum F=0$

In this way,

$Mg - T_1 = Ma$

$T_1 = Mg - Ma$ (1)

For block 2, we have the following relationship,

$T_2 = Ma$ (2)

So,

$T_1 - T_2 = I \frac{a}{R}$ (3)

Substituting (1) and (2) in (3)

$Mg - Ma - Ma = I \frac{a}{R}$

So,

$a = \frac{2\theta R}{t^2}$

**Note that $\alpha R = a$

c) For this point we need to use (1)

$T1 = Mg - Ma = Mg - \frac{M^2\theta R}{t^2}$

using (2)

$RT_2=RT_1-I\alpha$

$T_2= \frac{RM(g-\frac{2\theta R}{t^2})-I\frac{2\theta}{t^2}}{R}$

$T_2=\frac{RMgt^2-2R^2M\theta-2t\theta}{Rt^2}$