"I know not with what weapons World War III will be fought, but World War IV will be fought with sticks and stones."

– Albert Einstein

Questions on circular motion and rotation were discussed on many occasions on this site. You can access all those questions by clicking on the label ‘circular motion and rotation’ below this post. Or, you may try a search for*circular motion and rotation*using the search box provided on this page.

Today we will discuss a few more multiple choice practice questions in this section.

(1) A 7-metre long uniform iron girder of mass 300 kg is placed on the horizontal top of a building with 3 m extending over the edge of the building. A fireman weighing 75 kg starts walking towards the protruding end of the girder (Fig.). What is the maximum distance (from the edge of the building) upto which he can walk safely?

(a) 2 m

(b) 2.2 m

(c) 2.4 m

(d) 2.5 m

(e) 2.6 m

When the fireman moves just beyond the maximum safe distance, the girder starts rotating about the ege of the building. If the maximum safe istance is *d*, we have

75×*d* = 300×0.5

[We have equated the magnitude of the *moment of the weight of the fireman* (which is the deflecting torque due to the fireman’s weight) to the magnitude of the *moment of the weight of the girder* (which is the restoring torque due to the weight of the girder): 75*g*×*d* = 300*g*×0.5]

This gives *d = *2 m.

(2) A uniform metre stick of mass 0.2 kg can be balanced horizontally on a knife edge at the 40 cm mark, when a stone is suspended (using a string of negligible mass) from the 20 cm mark. The mass of the stone is

(a) 0.5 kg

(b) 0.3 kg

(c) 0.2 kg

(d) 0.1 kg

(e) 0.05 kg

The weight of the metre stick acts through its centre of gravity located at its middle (the 50 cm mark). Since metre stick is in equilibrium, the moment of the weight of the metre stick about the knife edge is equal and opposite to the moment of the weight of the stone about the knife edge. Therefore we have (on equating the magnitudes of the moments)

0.2*g*×0.1 = *Mg*×0.2 where *M* is the mass of the stone.

[Note that the distance of the knife edge from the centre of gravity of the metre stick is 0.1 m where as the distance from the line of action of the weight of the stone is 0.2 m].

(3) A square plate ABCD of side ‘*a*’* *and mass ‘*m’ *is suspended so that it can rotate about a horizontal axis passing through A and perpendicular to its plane. To rotate the plate, a horizontal force of magnitude *F* is applied to the plate at the corner C as shown in the figure. What is the magnitude of the torque produced by this force?

(a) *aF*

(b) *maF*

(c) (√2)*aF*

(d) *aF/*√2

(e) 2*maF*

The distance of the axis of rotation from the line of action of the force is √(*a*^{2}+*a*^{2}) = (√2)*a*. Therefore, the magnitude of the torque is (√2)*aF*.

[The mass of the plate (*m*) in the question just serves the purpose of distracting you!]

The following questions are meant for AP Physics C aspirants:

(4) A disc (of mass *M* and radius *R*) and a ring (of mass *M* and radius 2*R*) are released from rest from the top of an inclined plane. If they roll down the plane without slipping, which one of the following statements is correct?

(a) The ring will reach the bottom of the incline earlier

(b) They will reach the bottom of the incline with the same speed

(c) The ring will reach the bottom of the incline with twice the speed of the disc

(d) They will have the same acceleration down the incline

(e) They will reach the bottom of the incline with the same kinetic energy

Since the disc and the ring have the *same* mass and both have fallen down through the *same* height, they have lost the same amount of gravitational potential energy. Therefore, they have gained the *same* amount of kinetic energy so that the correct option is (e). Since the above question is a single correct answer type multiple choice question, you can easily arrive at the above answer.

[Let us consider the remaining options in the question (in view of the benefits you will have in certain situations).

The linear acceleration ‘*a*’* *of a body rolling down an inclined plane of angle *θ* is given by

*a = g *sin*θ/*[1*+*(*k*^{2}/*R*^{2})]* *where *g* is the gravitational acceleration, *k* is the radius of gyration about the axis of rolling and *R* is the radius of the rolling body.

In the case of a ring *k*^{2}/*R*^{2} = 1 where as in the case of a disc *k*^{2}/*R*^{2} = ½

(Remember that the moment of inertia of a body is *k*^{2} times the mass of the body).

The acceleration of the ring is less than that of the disc. In fact a ring (and a pipe) has the least acceleration while rolling down an incline. Options (a), (b), (c) and (d) are therefore wrong].

(5) A fly wheel has radius *R *and its axle has radius* R*/10. The ends of the axle are mounted on ball bearings so that the wheel can rotate without appreciable friction at the bearings. Strings are wound round the rim of the wheel and the axle. A mass *M* is suspended vertically from the free end of the string wound round the rim of the wheel. The free end of the string wound round the axle is pulled in a direction perpendicular to the length of the axle and making an angle* **θ* with the horizontal plane. If the mass *M* moves up with uniform velocity, what is the tension in the string wound round the axle?

(a) *Mg*

(b) *Mg*/10

(c) 10 *Mg*

(d) 10 *Mg *cos*θ*

(e) 10 *Mg*sin*θ*

The tension acting in the string wound round the rim of the wheel is *Mg* if the mass *M* is at rest or in uniform motion (moving with *uniform velocity*).

The torque produced by the weight of the suspended mas* M *is *RMg*. If the mass *M* is to be at rest or in uniform motion, the opposing torque to be applied by pulling on the string wound round the axle must be of the same magnitude *RMg*.

If *T *represents the tension in the string wound round the axle, we have

(*R*/10)×*T = RMg*

Therefore, *T* = 10*Mg *

[The angle *θ* in the question just serves the purpose of a distraction].

(6) A solid sphere of mass *M* and radius* R* rotates about its central axis with uniform angular acceleration α, on exerting a tangential force *F*_{1}. Another solid sphere of mass 2*M* and radius 2*R* rotates about its own central axis with the same angular acceleration α, on exerting a tangential force *F*_{2}. Then *F*_{1}* *and *F*_{2 }are related as

(a) *F*_{2} = 2*F*_{1}

(b) *F*_{2} = 4*F*_{1}

(c) *F*_{1} = 4*F*_{2}

(d) *F*_{1} = 2*F*_{2}

(e) *F*_{1} = *F*_{2}

The angular acceleration is the ratio of torque to the moment of inertia. Therefore, the angular accelerations of the spheres A and B are respectively *τ*_{1}/*I*_{1} and* τ*_{2}/*I*_{2} where *τ*_{1 }and *τ*_{2 }are the torques exerted on spheres A and B respectivly_{ }by the forces* F*_{1 }and *F*_{2} and* I*_{1} and *I*_{2 }are the moments of inertia of spheres A and B* *about their central axes.

Since the spheres have the same angular acceleration we have

*τ*_{1}/*I*_{1} = *τ*_{2}/*I*_{2}

Or, *R**F*_{1}/(2*MR*^{2}/5) = 2*R**F*_{2}/(2×2*M*×4*R*^{2}/5)

This gives *F*_{2} = 4*F*_{1}_{}

[Even if you don’t remember the expression for the moment of inertia of a solid sphere, you can work out this question if you know that the moment of inertia is directly proportional to the mass and the square of the radius].

You will find a few more multiple choice questions in this section here.

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