A disc of mass m1 is freely rotating with constant angular speed ω . Another disc of mass m2 & same radius is gently kept on the first disc. If the contacting surfaces are rough, the fractional decrease in kinetic energy of the system will be

Q: A disc of mass m1 is freely rotating with constant angular speed ω . Another disc of mass m2 & same radius is gently kept on the first disc. If the contacting surfaces are rough, the fractional decrease in kinetic energy of the system will be
(A)m1/m2
(B)m2/(m1+m2)
(C)m2/m1
(D) m1/(m1+ m2)

Solution: The upper disc speeds up and the lower disc slows down by the accelerating & rotating frictional retarding torques. However, the torque acting on the system is zero. Therefore, the angular momentum of the system remains constant.
 Linitial = Lfinal = L
rotation = 1-Iinitial/Ifinal
Iinitial = ½m1
Ifinal = ½m1r² + ½m2
B putting these values ,
ΔKE/KEinitial = m2/(m1+m2)

When a car negotiates a curve, the normal force exerted on the inner & outer wheels are N1 & N2 respectively . Then N1 / N2 is

Q: When a car negotiates a curve, the normal force exerted on the inner & outer wheels are N1 & N2 respectively . Then N1 / N2 is
(A) = 1
(B) < 1
(C) > 1
(D) zero

Solution:
rotation
Referring to the free body diagram we obtain,
ΣFy= N1 + N2 – mg = 0
N1 + N2 = mg . . . (A)
Along radial, ΣFr = mar
fr = mv2/r . . . (B)
ΣtG = 0
N1d/2 + frh – N2d/2 = 0    . . . (C)
Using N2 – N1 = (2h/d)(mv2/r) & N2 = mg/2 + hmv2/dr
N1/N2 < 1

A body of mass m and radius r is released from rest along a smooth inclined plane of angle of inclination θ . The angular momentum of the body about the instantaneous point of contact after a time t from the instant of release is equal to

Q: A body of mass m and radius r is released from rest along a smooth inclined plane of angle of inclination θ . The angular momentum of the body about the instantaneous point of contact after a time t from the instant of release is equal to
(A) mgrt cosθ
(B) mgrt sinθ
(C) (3/2) mgrt sinθ
(D) None of these

Solution : Since the surface is frictionless , the body does not rotate about its centre of mass. Only it slides with certain velocity v parallel to the surface
Hence,ω = 0   & v = (g sinθ) t
rotation
The angular momentum after a time t is given as
L = mvr
L = m(gsinθt) r
L = mgrt sinθ

A uniform disc of mass m and radius R is rolling down a rough inclined plane which makes an angle 30 degree with the horizontal. If the coefficients of static and kinetic friction are each equal to μ and the only forces acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is

Q: A uniform disc of mass m and radius R is rolling down a rough inclined plane which makes an angle 30º with the horizontal. If the coefficients of static and kinetic friction are each equal to μ and the only forces acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is
(A) (mg/3) upward
(B) (mg/3) downward
(C) (mg/6) upward
(D) (mg/6) downward

Solution :Laws of motion , Translational :
ma = mg sinθ  – f       . . . (i)
Rotation ,Torque  f.r = Iα      . . . (ii)
for rolling  α = a/r . . . (iii)    for disc, I =mR²/2
rotation
Solving (i) (ii) and (iii)
a =  g sinθ/(1+ I/mR²)
f = mgsinθ/(1+mR²/I)
= mgsin30°/(1+2)
= mg/6 Upward

Moment of inertia of a quarter disc having mass M and radius R about axis passing through centre of disc and perpendicular to plane is

Q: Moment of inertia of a quarter disc having mass m and radius R about axis passing through centre of disc and perpendicular to plane is
(A)I = mR²
(B)I = mR²/2
(C)I = mR²/4
(D)I = mR²/8

Solution: Moment of inertia (I) about z axis of a disc passing through centre of mass and perpendicular to the plane of disc is Iz = MR²/2
Since each quarter will have same moment of inertia (I)
4I= MR²/2
Iz = ¼MR²/2 = mR²/2
where ¼M = m = mass of each quarter of disc.