A wheel starts rotating from rest at time t=0 with a angular acceleration of 50 radians / s^2 . The angular acceleration (α) decreases to zero value after 5 seconds. During this interval, α varies according to the

Correct answer: C. 125 rad / s α=α_0 (1-(t)/(5)) At t=0, α=α_0 α_0=50 rad / s^2 (d ω)/(d t)=α_0(1-(t)/(5)) aligned & ∫_0^ω d ω=α_0 ∫_0^5(1-(t)/(5)) d t ω=α_0[t-(t^2)/(10)]_0^5 & =50(5-(25)/(10)) rad / s=125 rad / s aligned

AIIMS 2017

Exam: AIIMS
Year: 2017
Subject: Physics
Chapter: Motion in Two Dimensions

AIIMS 2017 · Physics · Motion in Two Dimensions

A wheel starts rotating from rest at time t=0 with a angular...

A wheel starts rotating from rest at time

t=0

with a angular acceleration of 50 radians

/ \mathrm{s}^2

. The angular acceleration

(\alpha)

decreases to zero value after 5 seconds. During this interval,

\alpha

varies according to the equation

\alpha=\alpha_0\left(1-\frac{t}{5}\right)

The angular velocity at

t=5 \mathrm{~s}

will be

Options

A $10 \mathrm{rad} / \mathrm{s}$
B $250 \mathrm{rad} / \mathrm{s}$
C $125 \mathrm{rad} / \mathrm{s}$
D $100 \mathrm{rad} / \mathrm{s}$

Correct answer

C. $125 \mathrm{rad} / \mathrm{s}$

Step-by-step solution

\alpha=\alpha_0

\left(1-\frac{t}{5}\right)

At

t=0, \alpha=\alpha_0 \quad \therefore \alpha_0=50 \mathrm{rad} / \mathrm{s}^2 \frac{d \omega}{d t}=\alpha_0\left(1-\frac{t}{5}\right)

\begin{aligned} & \therefore \int_0^\omega d \omega=\alpha_0 \int_0^5\left(1-\frac{t}{5}\right) d t \Rightarrow \omega=\alpha_0\left[t-\frac{t^2}{10}\right]_0^5 \\ & =50\left(5-\frac{25}{10}\right) \mathrm{rad} / \mathrm{s}=125 \mathrm{rad} / \mathrm{s}\end{aligned}

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