302 - Mechanics part 1 Questions Answers
Which of the following have maximum percentage of total K.E. in rotational form while pure rolling – Disc, Sphere, Ring or Hollow sphere. PLZ EXPLAIN ALSO.
total kinetic energy of a rolling body = mv2/2 + Iω2/2 = mv2/2 + Iv2/2r2 = (mv2/2)[1+I/mr2]
rotational kinetic energy of a rolling body = Iω2/2 = Iv2/2r2 = (mv2/2)[I/mr2]
so ratio R. K. E. / T. K. E. = (mv2/2)[I/mr2] / (mv2/2)[1+I/mr2] = [I/mr2] / [1+I/mr2]
= [I/I+mr2]
= 1/[1+(mr2/I)]
I is max. for a ring so ratio will be max. for a ring
To a billiard ball of mass M and radius r a horizontal impulse is given passing through its centre of mass and it provides velocity u to its centre of mass. If coefficient of friction between the ball and the horizontal surface is μ, then after how much time ball starts pure rolling?
for translational motion
0-μMg = Ma so a = -μg
so after time t velocity v = u--μgt
for rotational motion
about centre, μMgr = Iα = 2Mr2α/5
so α = 5μg/2r
so ω = 0 + αt
so ω = 5μgt/2r
for pure rolling
v = rω implies that u-μgt = r5μgt/2r
implies that u = 7μgt/2
so t = 2u/7μg
The thin circular ring shown below has mass M and length L. A force F acts at one end at an angle 30° with the horizontal and the rod is free to rotate about the other end in the plane of force. Initial angular acceleration of the rod is
ring or rod, first clear it
A thin circular ring slips down a smooth incline then rolls down a rough incline of identical geometry from same height. Ratio of time taken in the two motion is
in first slipping takes place whereas in second rolling takes place
for first accleration = g sinθ
and for second acceleration = g sinθ/[1+(I/mr2)]
now use s = ut + at2/2
A simple pendulum of mass and length L is held in horizontal position. If it is released from this position, then find the angular momentum of the bob about the point of suspension when it is vertically below the point of suspension.
by energy conservation v = √(2gL) below the point of suspension
so angular momentum = mvL about the point of suspension
A rod of length L and mass m is free to rotate about its one end in vertical plane. If it is released from horizontal position, then find torque on rod about end of rotation , when it makes an angle θ with vertical line
force is mg and perpendicular distance from axis l/2 sinθ
now calculate
Four solid rigid balls each of mass m and radius r are fixed on a rigid ring of radius 2r and mass 2m.The system is whirled about ‘O’. The radius of gyration of the system is 128 mr2/30. How???
about O
M. I. of ring = 2m(2r)2
M. I. of one ball = 2mr2/5 + m(2r)2 = 22mr2/5
so total = 8mr2 + 4*22mr2/5
solve
One other thing is that the question is wrong, in the place of radius of gyration, you should write moment of inertia
The rate of doing work by force acting on a particle moving along x-axis depends on position x of particle and is equal to 2x. The velocity of particle is given by
Rate of doing work is power so according to the given condition
P = 2x
so Fv = 2x
or m(dv/dt)v = 2x
or m(vdv/dx)v = 2x
or mv2dv = 2xdx
now integrate.
A shell at rest on a smooth horizontal surface explodes into two fragments of masses m1 and m2. If just after explosion m1 moves with speed u, then work done by internal forces during explosion is
By momentum conservation velocity of m2 = m1u/m2
now sum of kinetic energies = Net work done by internal forces
A body of mass m falls from height h on ground. If 'e' be the coefficient of restitution of collision between the body and ground, then the distance travelled by body before it comes to rest is
if a body falls from height h then height attained after rebound will be e2h
by using this fact
total distance covered before stop = h+2e2h+2e2e2h+2e2e2e2h+..................................................
so d = h+2e2h[1+e2+e4+e6+.........................]
now solve