# 42 - Rotational mechanics Questions Answers

Joshi sir comment

what is the moment of inertia of a cube about its diagonal?

Joshi sir comment

Joshi sir comment

A rotating ball hits a rough horizontal plane with a vertical velocity $v$ and angular velocity $w$. Given that the coefficient of friction is $u$ and the vertical component of the velocity after the collision is $v/2$, find:

a) the angular velocity after collision;

b) the impulsive ground reaction during the collision

Joshi sir comment

on applying momentum impulse theorem along vertical

-mv+Ndt = mv/2       (1)

by angular momentum angular impulse theorem

By eq. (2) get angular velocity.

impulsive ground reaction = $\sqrt{{\left(Ndt\right)}^{2}+{\left(\mu Ndt\right)}^{2}}$by using eq. (1), solve it

What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination θ?

Joshi sir comment

friction f = mgsinθ/[1+(mr2/I)]

solve for sphere and compare to μmgcosθ

Find the moment of inertia of a uniform square plate of mass M and edge a about one of its diagonals.

(1) Ma/ 12

(2) 2/3 x Ma2

Joshi sir comment

let the moment of inertia about a diagonal = I

then by perpendicular axis theorem 2I = 1/6 Ma2

now calculate I

Two sphere are rotating about their own axis if axis of rotation of these axis are perpendicular to each other than angular acceleration of one with respect to other
Joshi sir comment

let the angular velocities are ω1 ω2

then angular acceleration of one w. r. t. other =  ω1 ω2

A ball of mass 1 kg is projected with a velocity of 20√2 m/s from the origin of an xy coordinates axis system at an angle 45° with x-axis (horizontal). The angular momentum of the ball about the point of projection after 2s of projection is [take g=10 m/s2] (y-axis is taken as vertical)

Joshi sir comment

initial vertical and horizontal velocities are 20 m/s and 20 m/s

after 2 sec. vertical and horizontal velocities are 0 m/s and 20 m/s

and y and x displacements are 20 m and 40 m

so about point of projection angular momentum after 2 sec.

= mvxy - mvyx = 1*20*20 - 1*0*40 = 400

A disc of mass 3kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is

Joshi sir comment

mgh = 1/2 mv2 + 1/2 Iω2

I for the disc about centre = 1/2 MR

and for pure rolling v = Rω

calculate v then 1/2 mv2

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion

Joshi sir comment

for translational motion

0-μMg = Ma so a = -μg

so after time t velocity v = u-μgt  (1)

for rotational motion

about centre, μMgr = Iα = 2Mr2α/5

so α = 5μg/2r

so ω = 0 + αt

so ω = 5μgt/2r  (2)

for pure rolling

v = rω implies that u-μgt = r5μgt/2r

implies that u = 7μgt/2

so t = 2u/7μg

now put t in (1) for getting v