JEE Main Rotational Motion PYQ

21. (JEE Main 2021 (Online) 25th February Evening Shift)

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v₀. It encounters an inclined plane at angle $θ$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

JEE Main 2021 (Online) 25th February Evening Shift Physics - Rotational Motion Question 81 English

A. $\frac{v_{0}^{2}}{2 g \sin θ}$

B. $\frac{7 v_{0}^{2}}{10 g \sin θ}$

C. $\frac{2}{5} \frac{v_{0}^{2}}{g \sin θ}$

D. $\frac{v_{0}^{2}}{5 g \sin θ}$

Correct Answer is Option (B)

JEE Main 2021 (Online) 25th February Evening Shift Physics - Rotational Motion Question 81 English Explanation

From energy conservation

mgh = $\frac{1}{2} m v_{0}^{2} + \frac{1}{2} I ω^{2}$

mgh = $\frac{1}{2} m v_{0}^{2} + \frac{1}{2} \times \frac{1}{5} m a^{2} \times \frac{{v_{0}}^{2}}{a^{2}}$

gh = $\frac{1}{2} {v_{0}}^{2} + \frac{1}{5} {v_{0}}^{2}$

gh = $\frac{7}{10} {v_{0}}^{2}$

h = $\frac{7}{10} \frac{{v_{0}}^{2}}{g}$

from triangle, $\sin θ = \frac{h}{l}$

then h = l sin $θ$

$l \sin θ = \frac{7}{10} \frac{{v_{0}}^{2}}{g}$

$l = \frac{7}{10} \frac{{v_{0}}^{2}}{g \sin θ}$

22. (JEE Main 2021 (Online) 22th July Evening Shift )

The centre of a wheel rolling on a plane surface moves with a speed v₀. A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt{x} v_{0}$ . Then the value of x is _____________.

Correct answer is (02)

JEE Main 2021 (Online) 22th July Evening Shift Physics - Rotational Motion Question 62 English Explanation
$| ω | = \frac{v_{0}}{R}$

${\vec{v}}_{p} = v_{0} \hat{i} + ω R (- \hat{j}) = v_{0} \hat{i} - v_{0} \hat{j}$

$| {\vec{v}}_{p} | = \sqrt{2} v_{0}$

$x = 02$

23. (JEE Main 2021 (Online) 20th July Evening Shift )

Two bodies, a ring and a solid cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $\frac{\sqrt{x}}{2}$ . Then, the value of x is _____________.

Correct answer is (3)

I in both cases is about point of contact

Ring

mgh = $\frac{1}{2} I ω^{2}$

mgh = $= \frac{1}{2} (2 m R^{2}) \frac{v_{R}^{2}}{R^{2}}$

$v_{R} = \sqrt{g h}$

Solid cylinder

mgh = $\frac{1}{2} I ω^{2}$

mgh $= \frac{1}{2} (\frac{3}{2} m R^{2}) \frac{v_{C}^{2}}{R^{2}}$

$v_{C} = \sqrt{\frac{4 g h}{3}}$

$\frac{v_{R}}{v_{C}} = \frac{\sqrt{3}}{2}$

24. (JEE Main 2021 (Online) 20th July Morning Shift )

A circular disc reaches from top to bottom of an inclined plane of length 'L'. When it slips down the plane, it makes time 't₁'. When it rolls down the plane, it takes time t₂. The value of $\frac{t_{2}}{t_{1}}$ is $\sqrt{\frac{3}{x}}$ . The value of x will be _______________.

Correct answer is (2)

According to question, a circular disc reaches from top to bottom of an inclined plane of length L. This can be shown as

JEE Main 2021 (Online) 20th July Morning Shift Physics - Rotational Motion Question 66 English Explanation
When the disc slips down the inclined plane, it takes time t₁. Therefore, in this case its acceleration, a₁ = g sin $θ$

$∵$ s = ut₁ + $\frac{1}{2}$ a₁t $_{1}^{2}$

$\Rightarrow$ s = $\frac{1}{2}$ g sin $θ$ t $_{1}^{2}$ .... (i)

And when the disc rolls down the inclined plane, it takes time t₂. Therefore in this case, its acceleration,

$a_{2} = \frac{g \sin θ}{1 + \frac{K^{2}}{R^{2}}} = \frac{g \sin θ}{1 + \frac{1}{2}} = \frac{2}{3} g \sin θ$ [ $∵$ for disc, $\frac{K^{2}}{R^{2}} = \frac{1}{2}$ ]

$∵$ s = ut₂ + $\frac{1}{2}$ a₂t $_{2}^{2}$ = $\frac{1}{2}$ . $\frac{2}{3}$ g sin $θ$ t $_{2}^{2}$ .... (ii)

On dividing Eq. (i) by Eq. (ii), we get

$\frac{t_{2}}{t_{1}} = \sqrt{\frac{3}{2}}$ .... (iii)

According to question, value of $\frac{t_{2}}{t_{1}}$ is $\sqrt{\frac{3}{x}}$ .

Comparing it with Eq. (iii), we get x = 2

25. (JEE Main 2021 (Online) 17th March Morning Shift )

The following bodies,

(1) a ring

(2) a disc

(3) a solid cylinder

(4) a solid sphere,

of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]

JEE Main 2021 (Online) 17th March Morning Shift Physics - Rotational Motion Question 72 English

Correct answer is (4)

$a = \frac{g \sin θ}{(1 + \frac{I}{m R^{2}})}$

I_R = mR², a_R = g sin $θ$ /2

I_D = $\frac{m R^{2}}{2}$ , a_D = $\frac{2}{3}$ g sin $θ$

I_SC = $\frac{m R^{2}}{2}$ , a_SC = $\frac{2}{3}$ g sin $θ$

I_SC = $\frac{2}{5}$ mR², a_SS = $\frac{5}{7}$ g sin $θ$

S = ut + $\frac{1}{2}$ at²,

t = $\sqrt{\frac{2 S}{a}}$

$∴$ t $\propto$ $\frac{1}{\sqrt{a}}$

solid sphere will take minimum time.