68 - Rotational mechanics Questions Answers
What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination θ?
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) Ma2 / 12
(2) 2/3 x Ma2
let the moment of inertia about a diagonal = I
then by perpendicular axis theorem 2I = 1/6 Ma2
now calculate I
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)
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
mgh = 1/2 mv2 + 1/2 Iω2
I for the disc about centre = 1/2 MR2
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
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
Two rings of same mass and radius R are placed with their planes perpendicular to each other and centres at a common point. The radius of gyration of the system about an axis passing through the centre and perpendicular to the plane of one ring is
The axis passing through the centre of one ring and perpendicular to the plane will be diametric axis of the other ring
so moment of inertia about that axis = MR2 + 1/2 MR2 = 3/2 MR2
now compare it to Mk2 and get k, it is radius of gyration
A man of mass 60 kg is standing on a boat of mass 140 kg , which is at rest in still water. The man is initially at 20 m from the shore. He starts walking on the boat for 4s with constant speed 1.5 m/s towards the shore. The final distance of the man from the shore is
distance covered in boat = 1.5*4 = 6 m
and for making the position of centre of mass undistubed displacement of boat in opposite direction = x (let)
by law of conservation of momentum (140+60)*v = 60*1.5
so v = 90/200 = 0.45 m/s
so displacement of boat = 0.45*4 = 1.8 m
so final distance of the man from the shore = 20-6+1.8 = 15.8 m
A horizontal disc rotating freely about a vertical axis through its centre makes 90 revolutions per minute. A small piece of wax of mass m falls vertically on the disc and sticks to it at a distance r from the axis. If the number of revolutions per minute reduce to 60, then the moment of inertia of the disc is
use I1ω1 = I2ω2
let moment of inertia of disc about given axis is I
then I2π90/60 = [I+mr2]2π60/60
solve it
A uniform rod of mass m and length l is suspended by two strings at its ends. When one of the strings is cut, the rod starts falling with an initial angular acceleration
after cutting one string eq. of motion will be
mg-T = ma
mg*l/2 = Iα
put I = ml2/3 and get the answer