SATHEE: Rotational Motion Question 211

Rotational Motion Question 211

Question: The moment of inertia of a body about a given axis is $1.2 k g m^{2}$ . Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of $25 r a d i a n / \sec^{2}$ must be applied about that axis for a duration of

Options:

A) 4 seconds

B) 2 seconds

C) 8 seconds

D) 10 seconds

Show Answer

Answer:

Correct Answer: B

Solution:

[b]

$I = 1.2 kg m^{2}, E_{r} = 1500 J$

$α = 25 rad / \sec^{2}, ω_{1} = 0, t = ?$

$A s E_{r} = \frac{1}{2} I ω^{2}, ω = \sqrt{\frac{2 E_{r}}{I}}$ $= \sqrt{\frac{2 \times 1500}{1.2}} = 50 rad / \sec$

$From ω_{2} = ω_{1} + α t$

$50 = 0 + 25 t, t = 2 seconds$

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Rotational Motion Question 211

Question: The moment of inertia of a body about a given axis is1.2kgm2 . Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of25radian/sec2 must be applied about that axis for a duration of

Options:

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Question: The moment of inertia of a body about a given axis is $1.2 k g m^{2}$ . Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of $25 r a d i a n / \sec^{2}$ must be applied about that axis for a duration of