Question

A solid cylinder of mass \(\mathrm{M}\) and \(\mathrm{R}\) is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass \(\mathrm{m}\) is hung from the string as shown in figure the mass is released from rest then The angular speed of cylinder is proportional to \(\mathrm{h}^{\mathrm{n}}\), where \(\mathrm{h}\) is the height through which mass falls, Then the value of \(n\) is \(\\{\mathrm{A}\\}\) zero \(\\{\mathrm{B}\\} 1\) \(\\{\mathrm{C}\\}(1 / 2)\) \([\mathrm{D}] 2\)

Short answer

The value of \(n\) in the proportionality relationship between the angular speed of the solid cylinder and the height \(h\) is 1 (Option B).

Step-by-step solution

Identify the forces and torques

On this system, we have the following forces and torques: - Tension force on the string, T, acting upward on the mass "m" and acting tangentially on the cylinder. - Gravitational force on the falling mass, mg, acting downward. - Torque on the cylinder due to the tension force, τ = TR, with R being the radius of the cylinder, acting counterclockwise.

Apply Newton's second law and the relation for torque

For the hanging mass: \(m\frac{d^2h}{dt^2}=mg-T\) ... (1) For the cylinder: \(\frac{d^2\theta}{dt^2}=\frac{τ}{I}\) Where \(τ=TR\), and the moment of inertia of the solid cylinder \(I=\frac{1}{2}MR^2\). Therefore: \(\frac{1}{2}MR^2\frac{d^2\theta}{dt^2}=TR\) ... (2) Now, because the string doesn't slip, we have \(h=R\theta\), and taking the second derivative with respect to time, we get: \(\frac{d^2h}{dt^2}=R\frac{d^2\theta}{dt^2}\) Thus, from equations (1) and (2), we get: \(m\frac{d^2h}{dt^2}=mg-T\) \(mR\frac{d^2\theta}{dt^2}=TR\) Dividing these two equations, we get: \(\frac{m}{mR}=\frac{mg-T}{TR}\)

Determine the angular speed

From the previous equation, we can solve for the tension T: \(T=\frac{m^2g}{m+M}\) Now, we can find the angular acceleration \(\alpha=\frac{d^2\theta}{dt^2}\) by plugging the tension T back into equation (2): \(\frac{1}{2}MR^2\alpha=\left(\frac{m^2g}{m+M}\right)R\) Solving for \(\alpha\), we get: \(\alpha = \frac{2m^2g}{(m+M)MR}\) Using the kinematic equation for angular motion: \(\omega^2=\omega_0^2+2\alpha\theta\) Here, \(\omega\) is the final angular speed, \(\omega_0\) is the initial angular speed, and \(\theta\) is the angular displacement. Since the system starts from rest, \(\omega_0=0\), and as mentioned earlier, \(h=R\theta\). We can rewrite the equation: \(\omega^2=2\alpha h\) Now, plug in the value of \(\alpha\), which we derived earlier: \(\omega^2=2\frac{2m^2g}{(m+M)MR}h\)

Establish the proportionality relationship and find the value of "n"

Now, we can establish the proportionality relationship: \(\omega^2 \propto h^n\) Comparing this relationship with the equation for \(\omega^2\): \(\omega^2=2\frac{2m^2g}{(m+M)MR}h\) We find that the exponent "n" is: \(n=1\) Observing the choices: A) Zero B) 1 C) \(\frac{1}{2}\) D) 2 We can see that the correct answer is: \(\boxed{n = 1\ (Option\ B)}\)