System of Particles and Rotational Motion - Result Question 22
22. 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 $ms^{-2}$ at the end of $2.0 s$ is approximately:
[2016]
(a) 8.0
(b) 7.0
(c) 6.0
(d) 3.0
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Answer:
Correct Answer: 22. (a)
Solution:
- (a) Given: Radius of disc, $R=50 cm$
Angular acceleration, $\alpha=2.0 rads^{-2}$; time $t=2 s$
Particle at periphery (assume) will have both radial (one) and tangential acceleration
$a_t=R \alpha=0.5 \times 2=1 m / s^{2}$
Angular speed
$\omega=\omega_0+\alpha t$
$\omega=0+2 \times 2=4 rad / sec$
Centripetal acceleration
$a_c=\omega^{2} R=(4)^{2} \times 0.5=16 \times 0.5=8 m / s^{2}$
Linear acceleration, $a_t=\alpha \cdot R=1 m / sec^{2}$
Net acceleration,
$a _{\text{total }}=\sqrt{a_t^{2}+a_c^{2}}=\sqrt{1^{2}+8^{2}} \approx 8 m / s^{2}$
The acceleration of the particle moving in a circle has two components; tangential acceleration (at) which is along the tangent and radial acceleration (ar) which 15 towards the centre of circle. As the two components are mutually perpendicular therefore, net acceleration of the particle will be $a=\sqrt{a_t^{2}+a_r^{2}}$
