Question

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad $s^{-2}.$ Its net acceleration in $m{s}^{-2}$ at the end of 2.0 s is approximately :

• A. 3
• B. 8
• C. 7
• D. 6

Answer 1

The correct option is B 8

Given r = 50 cm = 50 $\times {10}^{-2}$ m
$\alpha$ = 2 rad $s^{-2}$
t = 2s
The tangential acceleration is $a_t=r\alpha =50\times {10}^{-2}\times 2=1m{s}^{-2}$
We know that $\omega ={\omega }_0+\alpha t$
$\omega =0+2\times 2=4rad{s}^{-1}$
The centripetal acceleration is $a_c=r{\omega }^2=50\times {10}^{-2}\times (4{)}^2=8m{s}^{-2}.$
So net acceleration is $a=\sqrt{a_t^2+a_c^2}=\sqrt{(1{)}^2+(8{)}^2}\approx 8m{s}^{-2}$