Rotational Motion

Practice Questions:

Question 1: difficult

An equilateral prism of mass m rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the prism as shown in the figure. If the coefficient of friction is sufficiently high so that the prism does not slide before toppling, then the minimum force required to topple the prism is :

1. mg/√3

2. mg/4

3. μmg/√3

4. 3mg/4

View Answer

Toppling will take place about right most point so,

F ×(√3a/2) = mg × a/2

⇒ F= mg/√3

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Question 2: difficult

A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its centre

1. m g a sin θ

2. (m g a sin θ)/3

3. (m g a sin θ)/4

4. (m g a sin θ)/2

View Answer

As the object slides with constant speed friction force is balancing mgsinθ. Torque of Normal reaction and friction force about centre of object will cancel each other.

τ=mg.sinθ × a/2 = (mg a sinθ)/2

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Question 3: difficult

A uniform pole of length l and mass m is pivoted on the ground with a frictionless hinge O. The pole is free to rotate without friction about an horizontal axis axis passing through O and normal to plane of the page. The pole makes an angle θ with the horizontal. The pole is released from rest in the position shown, then linear acceleration of the free end (P) of the pole just after its release would be

1. (2g cosθ)/3

2. 2g/3

3. g

4. (3g cosθ)/2

View Answer

Torque of gravitational force mg about O is mgcosθ ×L/2

Torque = I × α

mgcosθ ×L/2 = (mL²/3) × α

α = (3gcosθ)/2

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Question 4: difficult

A particle of mass 5g is moving with a uniform speed of 3 √2 cm/s in the x–y plane along the line y= 2 √5 cm. The magnitude of its angular momentum about the origin in g-cm²/s is

1. 0 (Zero)

2. 30

3. 30√2

4. 30√10

View Answer

The angular momentum

$L$

of a particle about the origin is given by:

$L = m v r \sin\theta$

where:

$m = 5$

g (mass of the particle),
$v = 3\sqrt{2}$

cm/s (speed of the particle),
$r = 2\sqrt{5}$

cm (perpendicular distance from the origin),
$\theta = 90^\circ$

(since the velocity is along a straight line parallel to the x-axis, the perpendicular distance is directly used).

Since

$\sin 90^\circ = 1$

, the equation simplifies to:

$L = m v r$

Substituting the given values:

$L = (5) \times (3\sqrt{2}) \times (2\sqrt{5})$

$L = 5 \times 3 \times 2 \times \sqrt{10}$

$L = 30 \sqrt{10} \text{ g-cm²/s}$

Thus, the magnitude of the angular momentum is:

$\mathbf{30\sqrt{10} \text{ g-cm²/s}}$

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