68 - Rotational mechanics Questions Answers
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
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