Circular Motion — Page 5 | JEE Physics

Q41

A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity

\omega

. The coefficient of static friction between the bottom of the beaker and the surface of the disc is

\mu

. The beaker will revolve with the disc if :

A

R \le {{\mu g} \over {2{\omega ^2}}}

B

R \le {{\mu g} \over {{\omega ^2}}}

C

R \ge {{\mu g} \over {2{\omega ^2}}}

D

R \ge {{\mu g} \over {{\omega ^2}}}

Correct Answer

Option B

Solution

The force that prevents the beaker from sliding is the static friction force.

For an object in uniform circular motion, the net force acting on the object (which is the friction force in this case) is equal to the centripetal force.

The static friction force is given by the normal force (which is equal to the weight of the object) multiplied by the coefficient of static friction, which is :

F_{\text{friction}} = \mu mg.

The centripetal force needed to keep an object moving in a circle of radius R at angular velocity ω is given by :

F_{\text{centripetal}} = mR\omega^2.

For the beaker not to slide off, the static friction force must be at least as large as the centripetal force.

Therefore, we have :

\mu mg \geq mR\omega^2.

After canceling the mass m from both sides, we get :

R \leq \frac{\mu g}{\omega^2}.

So, the correct answer is Option B :

R \leq \frac{\mu g}{\omega^2}.

Q42

A train is moving with a speed of

12 \mathrm{~m} / \mathrm{s}

on rails which are

1.5 \mathrm{~m}

apart. To negotiate a curve radius

400 \mathrm{~m}

, the height by which the outer rail should be raised with respect to the inner rail is (Given,

g=10 \mathrm{~m} / \mathrm{s}^2)

:

A 6.0 cm

B 5.4 cm

C 4.8 cm

D 4.2 cm

Correct Answer

Option B

Solution

\tan \theta=\frac{v^2}{R g}=\frac{12 \times 12}{10 \times 400}

\begin{aligned} & \tan \theta=\frac{\mathrm{h}}{1.5} \\\\ & \Rightarrow \frac{\mathrm{h}}{1.5}=\frac{144}{4000} \\\\ & \mathrm{~h}=5.4 \mathrm{~cm} \end{aligned}

Q43

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as

A zero

B mk2r2t2

C mk2r2t

D mk2rt

Correct Answer

Option C

Solution

{a_r} = {k^2}r{t^2} = {{{v^2}} \over r}

\Rightarrow {v^2} = {k^2}{r^2}{t^2}

or

v = krt

and

{{d|v|} \over {dt}} = kr

\Rightarrow {a_t} = kr

\Rightarrow |\overline F \,.\,\overline v | = (mkr)(krt)

= m{k^2}{r^2}t =

power delivered

Q44

A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be :

A 7.5 rad

B 15 rad

C 20 rad

D 30 rad

Correct Answer

Option B

Solution

{\theta _1} = {1 \over 2}\alpha (2 \times 1 - 1) = 5

rad $\Rightarrow$ $\alpha$ = 10 rad/sec2 So

{\theta _2} = {1 \over 2} \times \alpha (2 \times 2 - 1) = 15

rad

Q45

A vehicle of mass

200 \mathrm{~kg}

is moving along a levelled curved road of radius

70 \mathrm{~m}

with angular velocity of

0.2 ~\mathrm{rad} / \mathrm{s}

. The centripetal force acting on the vehicle is:

A

560 \mathrm{~N}

B

14 \mathrm{~N}

C

2800 \mathrm{~N}

D

2240 \mathrm{~N}

Correct Answer

Option A

Solution

The centripetal force acting on an object moving along a circular path of radius

r

and angular velocity $\omega$ is given by:

F_c=mr\omega^2

where

m

is the mass of the object. In this problem, the vehicle of mass

m=200 \mathrm{~kg}

is moving along a circular path of radius

r=70 \mathrm{~m}

with angular velocity

\omega=0.2 ~\mathrm{rad}/\mathrm{s}

. Substituting the given values into the formula, we get:

F_c=mr\omega^2=(200~\mathrm{kg})(70~\mathrm{m})(0.2~\mathrm{rad/s})^2=\boxed{560~\mathrm{N}}

Therefore, the centripetal force acting on the vehicle is

560~\mathrm{N}

.