302 - Mechanics part 1 Questions Answers
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
A particle of mass 5 kg is moving with a uniform speed 3√2 in XOY plane along Y = X + 4. The magnitude of its angular momentum about the origin is
perpendicular distance of the line x-y+4 = 0 from origin
= (0-0+4)/√2
so angular momentum = mvr = 5*(3√2)*4/√2 = 60
A wheel having moment of inertia 4 kg m2 about its symmetrical axis, rotates at rate of 240 rpm about it. The torque which can stop the rotation of the wheel in one minute is
angular retardation = 2π[240/60]/60 this is by the formula ω = ω0+αt where ω0 = 0
now use τ = Iα
Two equal and opposite forces are applied tangentially to a uniform disc of mass M and radius R as shown in the figure. If the disc is pivoted at its centre and free to rotate in its plane, the angular acceleration of the disc is
by using τ = Iα
2FR = MR2α/2
solve?
A disc of mass M kg and radius R metre is rotating at an angular speed of ω rad/s when the motor is switched off. Neglecting the friction at the axle, the force that must be applied tangentially to the wheel to bring it to rest in time t is
let force is F
then retarding torque about axis = FR
so by τ = Iα calculate α
then use first eq. of motion in rotation
ωf = ωi - αt
Four thin uniform rods each of length l and mass m are joined to form a square. The moment of inertia of square about an axis along its one diagonal is
moment of inertia of 1 rod about diameter = m(lsin45)2/3
now multiply it to 4 for getting the net inertia about diameter
A disc of mass 1 kg and radius 0.1m is rotating with angular velocity 20 rad/s. What is angular velocity (in rad/s) if a mass of 0.5 kg is put on the circumference of the disc?
by angular momentum conservation
I1ω1 = I2ω2
so [1(0.1)2/2]20 = [1(0.1)2/2+0.5(0.1)2]ω2
solve and get ω2