# Answer to Question #123984 in Mechanics | Relativity for Abhi

Question #123984
Awheel rotating about a fixed axis at 24 r.p.m. is uniformly accelerated for 70 seconds,
during which time it makes 50 revolution. Find (i) angular velocity at the end of this
interval, and (ii) time required for the speed to reach 150 r.p.m.
(Ans. - 61.4 r.p.m. ;3 min 55.6 s)
1
2020-06-26T14:36:30-0400

Explanations & Calculations

• Equivalent forms of the equations used in linear motion under uniform acceleration are used in rotational motion.

1).

• Apply "\\small \\theta = \\Big(\\frac{\\omega_{final}+\\omega_{initial}}{2} \\Big)\\small t" to calculate the angular velocity.

"\\qquad\\qquad\n\\begin{aligned}\n\\small 50 rev. &= \\small \\Big(\\frac{\\omega_{final}+24 \\,r.p.m}{2}\\Big)\\times \\Big(\\frac{70}{60} min\\Big)\\\\\n\\small \\omega_{final}&= \\small \\bold{61.71 \\,rpm }\n\\end{aligned}"

• Before stepping forward, angular acceleration ("\\alpha" )should be calculated. Apply "\\small \\omega_{final} = \\omega_{initial} +\\alpha t" to calculate it.

• And there is no need to calculate it as a final figure since it's not asked to calculate.

2)

• Apply again the above relationship from start to when it makes 150 rpm s.

• By (1) = (2), time take to reach 150 rpm s,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\frac{61.71 -24}{(\\frac{70}{60})} &= \\small \\frac{150 rpm -24rpm}{t} \\\\\n\\small t &= \\small 3.898min\\cdots(3min+0.898min)\\\\\n&= \\small 3min +(0.898\\times60s)\\\\\n&= \\small \\bold{3min\\,53.9s}\n\\end{aligned}"

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Assignment Expert
29.06.20, 21:23

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Abhi
28.06.20, 07:04

A stone is dropped from the top of a cliff 120 metres high. After one second, another stone is thrown down and strikes the first stone when it has just reached the foot of the cliff. Find the velocity with which the second stone was thrown.