Tag Archives: Rotational Mechanics

Consider a 20 kg uniform circular disk …

Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its centre and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around its periphery as shown in the figure. Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad/s. The value of n, to the nearest integer, is ———. [Given : In one complete revolution, the disk rotates by 6.28 rad] Continue reading Consider a 20 kg uniform circular disk …

A horizontal force F is applied at the center of mass …

A horizontal force F is applied at the center of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ. The center of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

(A) For the same F, the value of a does not depend on whether the cylinder is solid or hollow
(B) For a solid cylinder, the maximum possible value of a is 2μg
(C) The magnitude of the frictional force on the object due to the ground is always μmg
(D) For a thin-walled hollow cylinder, $𝑎 = \frac {𝐹}{2𝑚}$ Continue reading A horizontal force F is applied at the center of mass …

A thin rod of mass M and length a ….

A thin rod of mass M and length a is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius a/4 is pivoted on this rod with its center at a distance a/4 from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\Omega$. The total angular momentum of the system about the point O is $(\frac {Ma^2\Omega}{48}) n$.

The value of n is ___.

Continue reading A thin rod of mass M and length a ….

A 2 kg steel rod of length 0.6 m is clamped ….

A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is _ _ _ _ $m s^{-1}$.

Solution

Decrease in gravitational potential energy = Increase in kinetic energy

So, $mgl = \frac {1}{2} I \omega ^2 = \frac {1}{2} .\frac {1}{3} ml^2 \omega ^2 $

$\Rightarrow 6g = l \omega ^2 $

Using $v = l \omega $, we have

$6 g = l. \frac {v^2}{l^2}$

$\Rightarrow v^2 = 6gl $

$\Rightarrow v = \sqrt {6gl} = \sqrt {6\times 10 \times 0.6 } = 6ms^{-1} $

Answer: 6