Answer

Verified

388.8k+ views

**Hint:**As a first step, you could make a free body diagram marking all the forces. Then you could apply an equation of motion for both rotational and sliding motion. You could then make necessary changes in these relations and then equate them each other accordingly in each case. Finally, you could get two relations of linear initial velocity with initial angular velocity.

**Formula used:**

Equation of motion,

$v={{v}_{0}}+at$

Moment of inertia of marble,

$I=\dfrac{2}{5}m{{r}^{2}}$

**Complete answer:**

In the question, we have a marble that is being squeezed out from under the finger with linear speed ${{v}_{0}}$ and a simultaneous rotational speed of${{\omega }_{0}}$. Let $\mu $ be the coefficient of sliding friction. We are also given two cases under this information.

Firstly, let us consider the free body diagram of the marble and then mark all the forces that are acting on it.

The force that is responsible for the sliding and rolling motion is frictional force.

From Newton’s equation of motion we have,

$v={{v}_{0}}+at$

But,

$F=ma=\mu mg$

$\Rightarrow a=\mu g$

$\therefore v={{v}_{0}}-\mu gt$ ………………………………………. (1)

Now for the rolling motion we have,

$\omega ={{\omega }_{0}}+\alpha t$

Where, $\alpha $ is the rotational acceleration.

We know that torque is given by,

$\tau =I\alpha =Fr$

$\Rightarrow \alpha =\dfrac{Fr}{I}=\dfrac{\mu mgr}{I}$

$\therefore \omega ={{\omega }_{0}}-\dfrac{\mu mgr}{I}t$ ……………………………………… (2)

(1) Now for the first case, final linear and angular velocity is zero. So equations (1) and (2) becomes,

$0={{v}_{0}}-\mu gt$

$\Rightarrow t=\dfrac{{{v}_{0}}}{\mu g}$ ………………………. (3)

$0={{\omega }_{0}}-\dfrac{\mu mgr}{I}t$

$\Rightarrow t=\dfrac{I{{\omega }_{0}}}{\mu mgr}$ …………………… (4)

From (3) and (4),

$\dfrac{{{v}_{0}}}{{{\omega }_{0}}}=\dfrac{I}{mr}$

Now for the moment of inertia marble,

$I=\dfrac{2}{5}m{{r}^{2}}$

$\Rightarrow \dfrac{{{v}_{0}}}{{{\omega }_{0}}}=\dfrac{\left( \dfrac{2}{5}m{{r}^{2}} \right)}{mr}$

$\therefore {{v}_{0}}=\dfrac{2}{5}r{{\omega }_{0}}$ …………………………………… (5)

(2) Now for the second case,$v=0$ and $\omega =\dfrac{{{\omega }_{0}}}{2}$

$\Rightarrow \dfrac{{{v}_{0}}}{\mu g}=\dfrac{I{{\omega }_{0}}}{2\mu mgr}$

$\Rightarrow {{v}_{0}}=\dfrac{\left( \dfrac{2}{5}m{{r}^{2}} \right)}{2mr}{{\omega }_{0}}$

$\therefore {{v}_{0}}=\dfrac{1}{5}r{{\omega }_{0}}$ ………………………………………………. (6)

**Comparing equations (5) and (6) along with the given options in the question, we find that options B and C are the true conditions.**

**Note:**

We are clearly given in the question that the body is going to rest after sliding in both the cases. So clearly we could conclude that the marble is undergoing decelerated motion. Therefore, we have assigned the sign for acceleration as negative as the velocity is being reduced over time.

Recently Updated Pages

What number is 20 of 400 class 8 maths CBSE

Which one of the following numbers is completely divisible class 8 maths CBSE

What number is 78 of 50 A 32 B 35 C 36 D 39 E 41 class 8 maths CBSE

How many integers are there between 10 and 2 and how class 8 maths CBSE

The 3 is what percent of 12 class 8 maths CBSE

Find the circumference of the circle having radius class 8 maths CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One cusec is equal to how many liters class 8 maths CBSE