A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v. It collides elastically and head on with an identical sphere Bat rest. Neglect friction everywhere. After the collision, their angular speeds are ω and ω_B, respectively. Then

ω_A < ω_B
ω_A = ω_B
ω_A = ω
ω_B < ω

Solution

Since the spheres are smooth,there will be no transfer of angular momentum from the sphere A to sphere B.The sphere A only transfers its linear velocity v to the sphere B and will continue to rotate with the same angular speed ω.

A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin

is zero
remains constant
goes on increasing
goes on decreasing.

Solution

Angular momentum of mass m moving with a constant velocity about origin is

L = momentum × perpendicular distance of line of action of momentum from origin

L = mv × y

In the given condition mvy is a constant. There fore angular momentum is constant.

A couple is acting on a two particle systems. The resultant motion will be

purely rotational motion
purely linear motion
both a and b
None of these

Solution

A couple consists of two equal and opposite forces whose lines of actionare parallel and laterally separated by same distance. Therefore, net force (or resultant) of a couple is null vector, hence no translatory motion will be produced and only rotational motion will be produced.

A mass is tied to a string and rotated in a vertical circle, the minimum velocity of the body at the top is

\(\sqrt{gr}\)
g /r
\(\left ( \frac{g}{r} \right )^{3/2}\)
gr

Solution

A particles performing uniform circular motion.Its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is

^L⁄₄
2 L
4 L
^L⁄₂

Solution

A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness ^t⁄₄. Then the relation between the moment of inertia I_X and I_Y is

I_Y = 32 I_X
I_Y = 16 I_X
I_Y = I_X
I_Y = 64 I_X

Solution

A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T₁ and T₂ be the tensions at the points L/4 and 3L/4 away from the pivoted end. Then

T₁ > T₂
T₂ > T₁
T₁– T₂
the relation between T₁ and T₂ depends on whether the rod rotates clockwise and anticlockwise

Solution

Tension provides the necessary centripetal force for the rest of rod.

The ratio of moment of inertia of circular ring & circular disc having the same mass & radii about on axis passing the c.m & perpendicular to plane is

1 : 1
2 : 1
1 : 2
4 : 1

Solution

Moment of inertia (M.I) of circular ring I₁ = mr²

moment of inertia (M.I) of circular disc I₂ = ½ mr²

⇒ ^I₁⁄_I₂ = ²⁄_I

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