SATHEE: Rotational Motion Question 217

Rotational Motion Question 217

Question: A tangential force of 20 N is applied on a cylinder of mass 4 kg and moment ofinertia $0.02 k g m^{2}$ about its own axis. If the cylinder rolls without slipping, then linear acceleration of its centre of mass will be

Options:

A) $6.7 m / s^{2}$

B) $10 m / s^{2}$

C) $3.3 m / s^{2}$

D)None of these

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Let friction force = f $F + f = m a$ - (i) $(F - f) R = I α$ - (ii) $

From $e q^{n} s .$ (i) and (ii),

$2 F = m a + \frac{I α}{R}$

$U s e α = \frac{a}{R}$ (for pure rolling)

$2 F = m a + \frac{I a}{R^{2}}$ $40 = 4 a + \frac{0.02 a}{{(0.1)}^{2}};$

$a = \frac{40}{6} = 6.7 m / s^{2}$

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Rotational Motion Question 217

Question: A tangential force of 20 N is applied on a cylinder of mass 4 kg and moment ofinertia 0.02kgm2 about its own axis. If the cylinder rolls without slipping, then linear acceleration of its centre of mass will be

Options:

Answer:

Solution:

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Question: A tangential force of 20 N is applied on a cylinder of mass 4 kg and moment ofinertia $0.02 k g m^{2}$ about its own axis. If the cylinder rolls without slipping, then linear acceleration of its centre of mass will be