SATHEE: System of Particles and Rotational Motion

System of Particles and Rotational Motion - Result Question 80

86. A solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping. The speed of its centre of mass when it reaches the bottom is

[2003, 1989]

(a) $\sqrt{(2 g h)}$

(b) $\sqrt{4 g h / 3}$

(d) $\sqrt{4 g / h}$

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Answer:

Correct Answer: 86. (b)

Solution:

$\begin{aligned} (b) & K . E . = \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2} \\ K . E . & = \frac{1}{2} (\frac{1}{2} m r^{2}) ω^{2} + \frac{1}{2} m v^{2} \\ = \frac{1}{4} m v^{2} + \frac{1}{2} m v^{2} = \frac{3}{4} m v^{2} \end{aligned}$

Now, gain in K.E. $=$ Loss in P.E.

$\begin{matrix} (b) & \frac{3}{4} m v^{2} = m g h \Rightarrow v = \sqrt{(\frac{4}{3}) g h} \end{matrix}$

Torque, $I α = f . R$ .

Using Newton’s IInd law, $m g \sin θ - f = m a$

$∵$ pure rolling is there, $a = R α$

$\Rightarrow m g \sin θ - \frac{I α}{R} = m a$

$\Rightarrow m g \sin θ - \frac{I a}{R^{2}} = m a (∵ α = \frac{a}{R})$

or, acceleration, $a = \frac{m g \sin θ}{(I / R^{2} + m)}$

Using, $s = u t + \frac{1}{2} a t^{2}$

or, $s = \frac{1}{2} a t^{2} \Rightarrow t α \frac{1}{\sqrt{a}}$

$t$ minimum means a should be more. This is possible when $I$ is minimum which is the case for solid cylinder.

Therefore, solid cylinder will reach the bottom first.

System of Particles and Rotational Motion - Result Question 80

86. A solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping. The speed of its centre of mass when it reaches the bottom is

Answer:

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System of Particles and Rotational Motion - Result Question 80

86. A solid cylinder of mass m and radius R rolls down an inclined plane of height h without slipping. The speed of its centre of mass when it reaches the bottom is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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86. A solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping. The speed of its centre of mass when it reaches the bottom is