JEE Main 2021 (27 July Shift-II)

LEVELJEE Main

Animated Solution for Physics - Rotational Motion: In the given figure, two wheels $P$ and $Q$ are connected by a belt $B$ . The radius of $P$ is three times as that of $Q$ . In case of same rotational kinetic energy, the ratio of rotational inertias $(\frac{I _{1}}{I _{2}})$ will be $x : 1$ . The value of $x$ will be ………… .

Q: What if the belt was slipping?

If the belt slips, the no-slip condition $v = \omega R$ would not hold true for both wheels simultaneously. The problem would require additional information about the friction and slipping rate to solve.

Enter Numerical Value:

Visualized Solution

$v_{P} = v_{Q} = v$

Let the linear speed of the belt be $v$ .
Since the belt does not slip, the tangential speed at the rim of both wheels is $v$ .

$ω = \frac{v}{r}$

Relation between linear and angular speed: $ω = \frac{v}{r}$
Rotational Kinetic Energy: $K E = \frac{1}{2} I ω^{2}$

$K E_{P} = K E_{Q}$

Given, $K E_{P} = K E_{Q} ⟹ \frac{1}{2} I_{1} ω_{1}^{2} = \frac{1}{2} I_{2} ω_{2}^{2}$
Substituting $ω_{1} = \frac{v}{3 R}$ and $ω_{2} = \frac{v}{R}$

$\frac{1}{2} I_{1} (\frac{v}{3 R})^{2} = \frac{1}{2} I_{2} (\frac{v}{R})^{2}$

$\frac{1}{2} I_{1} (\frac{v}{3 R})^{2} = \frac{1}{2} I_{2} (\frac{v}{R})^{2}$
$⟹ I_{1} \frac{v ^{2}}{9 R ^{2}} = I_{2} \frac{v ^{2}}{R ^{2}}$

$\frac{I _{1}}{I _{2}} = 9$

Cancelling common terms: $\frac{I _{1}}{9} = I_{2}$
$⟹ \frac{I _{1}}{I _{2}} = \frac{9}{1}$
Comparing with $\frac{x}{1}$ , we get $x = 9$

$Conclusion$

In gear systems, $v$ is constant at the contact point.
A larger gear has more inertia but rotates slower, maintaining energy balance.

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The Sigma Insight: Work and Energy in Rotational Motion

Solution Diagram

Analyzing the Setup Imagine two wheels, $P$ and $Q$ , connected by a tight belt $B$

The radius of wheel

P

3 R

, which is three times the radius of wheel

Q

(

R

Because the belt connects the rims of both wheels and we assume it doesn't slip, the tangential (linear) speed at the rim of both wheels must be exactly the same. Let's call this common linear speed

v

The Master Equation We are given that both wheels possess the exact same rotational kinetic energy

The formula for rotational kinetic energy is:

K E = \frac{1}{2} I ω^{2}

where

I

is the moment of inertia and

ω

is the angular speed.

We also know the fundamental relationship between linear speed

v

and angular speed

ω

for a point on the rim of a rotating body:

v = ω r ⟹ ω = \frac{v}{r}

Let's express the angular speeds of both wheels in terms of the common linear speed

v

: For wheel

P

ω_{1} = \frac{v}{3 R}

For wheel

Q

ω_{2} = \frac{v}{R}

Final Calculation

Now, we equate their rotational kinetic energies:

K E_{P} = K E_{Q}

\frac{1}{2} I_{1} ω_{1}^{2} = \frac{1}{2} I_{2} ω_{2}^{2}

Substitute the expressions for

ω_{1}

and

ω_{2}

\frac{1}{2} I_{1} (\frac{v}{3 R})^{2} = \frac{1}{2} I_{2} (\frac{v}{R})^{2}

Don't rush through this. Let's carefully square the terms inside the brackets:

I_{1} \frac{v ^{2}}{9 R ^{2}} = I_{2} \frac{v ^{2}}{R ^{2}}

Notice how the

v^{2}

and

R^{2}

terms are present on both sides. We can elegantly cancel them out, along with the

\frac{1}{2}

\frac{I _{1}}{9} = I_{2}

Rearranging this to find the ratio of their rotational inertias:

\frac{I _{1}}{I _{2}} = \frac{9}{1}

The problem states this ratio is

x : 1

. Comparing the two, we get our final answer: $x = 9$

Common Traps & FAQs

Why is the linear speed

v

the same for both wheels?

Because they are connected by an inextensible belt that doesn't slip. If one wheel's rim moved faster than the other, the belt would either stretch or bunch up, which violates the rigid belt assumption.

Does a larger radius always mean a larger moment of inertia?

Not necessarily. Moment of inertia depends on both mass and how that mass is distributed (

I = \int r^{2} d m

). However, in this specific problem, we are finding the ratio based purely on the energy constraint, regardless of their actual masses.

What if the belt was slipping?

If the belt slips, the no-slip condition

v = ω R

would not hold true for both wheels simultaneously. The problem would require additional information about the friction and slipping rate to solve.

Why do we use

\frac{1}{2} I ω^{2}

instead of

\frac{1}{2} m v^{2}

The wheels are rotating about fixed central axes. They don't have any translational motion (their centers of mass aren't moving). Therefore, their total kinetic energy is purely rotational.

Can we find the individual moments of inertia from this data?

No, we can only find their ratio. To find the exact values of

I_{1}

and

I_{2}

, we would need the actual value of the kinetic energy and the linear speed

v

Enter Numerical Value:

Visualized Solution

$v_{P} = v_{Q} = v$

$ω = \frac{v}{r}$

$K E_{P} = K E_{Q}$

$\frac{1}{2} I_{1} (\frac{v}{3 R})^{2} = \frac{1}{2} I_{2} (\frac{v}{R})^{2}$

$\frac{I _{1}}{I _{2}} = 9$

$Conclusion$

The Sigma Insight: Work and Energy in Rotational Motion

Analyzing the Setup Imagine two wheels, $P$ and $Q$ , connected by a tight belt $B$

The Master Equation We are given that both wheels possess the exact same rotational kinetic energy

Final Calculation

Similar Questions

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One end of a straight uniform 1 m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30º from the horizontal (see figure). Its angular speed when its hits the table is given as $n s^{- 1}$ , where $n$ is an integer. The value of $n$ is ......

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vP​=vQ​=v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=rv​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } KEP​=KEQ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 21​I1​(3Rv​)2=21​I2​(Rv​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​I1​​=9

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

The Sigma Insight: Work and Energy in Rotational Motion

The Master Equation We are given that both wheels possess the exact same rotational kinetic energy

Final Calculation

Similar Questions

$v_{P} = v_{Q} = v$

$ω = \frac{v}{r}$

$K E_{P} = K E_{Q}$

$\frac{1}{2} I_{1} (\frac{v}{3 R})^{2} = \frac{1}{2} I_{2} (\frac{v}{R})^{2}$

$\frac{I _{1}}{I _{2}} = 9$

$Conclusion$