Question

Single Correct Answer Type
The angular velocity of a particle moving in a circle of radius $50$ cm is increased in $5$ min from $100$ revolutions per minute to $400$ revolutions per minute. Find the tangential acceleration of the particle.

• A. $60m{s}^{-2}$
• B. $\pi /30m{s}^{-2}$
• C. $\pi /15m{s}^{-2}$
• D. $\pi /60m{s}^{-2}$

Answer 1

Answer: (D)

$\pi /60m{s}^{-2}$

Initial angular velocity of particle is ${\omega }_1=\frac{200\pi }{60}rad{s}^{-1}$

Final angular velocity of particle is ${\omega }_2=\frac{800\pi }{60}rad{s}^{-1}$

angular acceleration $\alpha =$$\frac{{\omega }_2-{\omega }_1}{T}$

where T is total time which is equal to $5\times 60=300sec$

$\Rightarrow \alpha =\frac{\frac{800\pi }{60}-\frac{200\pi }{60}}{300}=\frac{10\pi }{300}=\frac{\pi }{30}rad{s}^{-2}$

Tangential acceleration of the particle $a=\alpha r$

where r is radius of the circle, $r=\frac{1}{2}m$

$\Rightarrow a=\frac{\pi }{30}\times \frac{1}{2}=\frac{\pi }{60}m{s}^{-2}$

correct answer is D.