Question

The angular acceleration of a particle moving along a circle is given by $\alpha = k \sin \theta$, where $\theta$ is the angle turned by particle and $k$ is constant. The centripetal acceleration of the particle in terms of $k$, $\theta$ and $R$ (radius of circle) is given by :

• A. $2kR(\sin^2 \theta)$
• B. $2kR(\sin \theta)$
• C. $2kR(1 + \cos \theta)$
• D. $2kR(1 - \cos \theta)$

Answer 1

Answer: (D)

2kR (1 - cos θ)

Answer 2

The angular acceleration $\alpha$ is given by $\alpha = k \sin \theta$. We can relate angular acceleration to angular velocity using the chain rule:

$\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta} \frac{d\theta}{dt} = \omega \frac{d\omega}{d\theta}$.

Therefore, $\omega \frac{d\omega}{d\theta} = k \sin \theta$.

Integrating both sides, assuming the particle starts from rest (i.e., at $\theta = 0$, $\omega = 0$):

$\int_0^\omega \omega' d\omega' = \int_0^\theta k \sin \theta' d\theta'$

$\frac{\omega^2}{2} = k (1 - \cos \theta)$

$\omega^2 = 2k (1 - \cos \theta)$

The centripetal acceleration $a_c$ is given by $a_c = R\omega^2$.

Substituting the expression for $\omega^2$:

$a_c = R [2k (1 - \cos \theta)] = 2kR (1 - \cos \theta)$