Question

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration $a_c$ is varying with time as $a_c = k^2rt^2$, where k is a constant. The power delivered to the particle by the forces acting on it is:

• A. $2\pi mk^2r^2t$
• B. $mk^2r^2t$
• C. $mk^4r^2t^5/3$
• D. 0

Answer 1

Answer: (B)

$mk^2r^2t$

Answer 2

The power delivered to a particle is given by the dot product of the net force acting on it and its velocity, $P = \vec{F} \cdot \vec{v}$.

In circular motion, the net force can be resolved into two components:

1. Centripetal force ($F_c$): This force acts radially inwards and is responsible for changing the direction of velocity. It is always perpendicular to the velocity vector. Therefore, the power delivered by the centripetal force is $F_c \cdot \vec{v} = 0$.

2. Tangential force ($F_t$): This force acts along the tangent to the circular path and is responsible for changing the magnitude of the velocity (speed). It is always parallel to the velocity vector. Therefore, the power delivered by the tangential force is $P = F_t v$.

Given:

• Mass of the particle = $m$
• Radius of the circular path = $r$ (constant)
• Centripetal acceleration $a_c = k^2rt^2$

We know that centripetal acceleration is also given by $a_c = \frac{v^2}{r}$, where $v$ is the speed of the particle.

Equating the two expressions for centripetal acceleration: $k^2rt^2 = \frac{v^2}{r}$

Solving for $v^2$: $v^2 = k^2r^2t^2$

Taking the square root to find the speed $v$: $v = \sqrt{k^2r^2t^2} = krt$ (assuming $k, r, t$ are positive, which they are for physical quantities).

Now, we need to find the tangential acceleration $a_t$. Tangential acceleration is the rate of change of speed: $a_t = \frac{dv}{dt}$

Substitute the expression for $v$: $a_t = \frac{d}{dt}(krt)$ Since $k$ and $r$ are constants: $a_t = kr \frac{d}{dt}(t)$ $a_t = kr \times 1 = kr$

The tangential force $F_t$ is given by Newton's second law: $F_t = m a_t$ $F_t = m(kr) = mkr$

Finally, the power delivered to the particle is the product of the tangential force and the speed: $P = F_t v$ $P = (mkr)(krt)$ $P = mk^2r^2t$