Problem 6 - Master Resource Book in JEE Main Physics

Question

A particle of mass \(m\) is moving in circular path of constant radius \(r\) such that its centripetal acceleration \(a_{c}\) is varying with time \(t\) as \(a_{c}=k^{2} r t^{2}\). The power delivered to the particle by the forces acting on it is (a) \(2 \pi m k^{2} r^{2} t\) (b) \(m k^{2} r^{2} t\) (c) \(\frac{m k^{4} r^{2} t^{5}}{3}\) (d) zero

Step-by-Step Solution

Answer

The power delivered is \(m k^2 r^2 t\), matching option (b).

Step 1: Express acceleration in terms of velocity

The centripetal acceleration \(a_{c}\) is related to the velocity \(v\) of the particle and the radius \(r\) of the path by the formula \(a_{c} = \frac{v^2}{r}\). Given \(a_{c} = k^2 r t^2\), we have: \[\frac{v^2}{r} = k^2 r t^2 \]Solving for \(v^2\), we find:\[v^2 = k^2 r^2 t^2\]

Step 2: Determine the expression for velocity

By taking the square root of both sides, we determine the velocity \(v\):\[ v = k r t \]

Step 3: Compute the instantaneous power delivered

The instantaneous power \(P\) delivered to the particle is given by the rate of change of kinetic energy \(KE = \frac{1}{2} m v^2\). The expression for power is:\[P = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right)\]Substitute \(v^2 = k^2 r^2 t^2\):\[KE = \frac{1}{2} m (k^2 r^2 t^2)\]Differentiate \(KE\) with respect to \(t\):\[P = \frac{d}{dt} \left( \frac{1}{2} m k^2 r^2 t^2 \right) = m k^2 r^2 t\]

Step 4: Verify the solution options

Compare the derived expression for power \(P = m k^2 r^2 t\) to the provided answer choices. - Option (b) matches exactly with our derived expression: \(m k^2 r^2 t\). Thus, option (b) is correct.

Key Concepts

Centripetal Acceleration

When a particle moves along a circular path, it constantly changes direction. This change requires a special kind of acceleration called "centripetal acceleration." Imagine swinging a ball tied to a string around in a circle. The string pulls the ball towards the center, creating this acceleration. The term "centripetal" literally means "center-seeking." This force does not speed up or slow down the ball; rather, it keeps it moving in a circular path.

The formula for centripetal acceleration is given by \[ a_{c} = \frac{v^2}{r} \]where:

• \( a_{c} \) is the centripetal acceleration.
• \( v \) is the velocity of the particle.
• \( r \) is the radius of the circle.

For this exercise, the centripetal acceleration changes over time: \[ a_{c} = k^{2} r t^{2} \]This tells us that not only is the particle's direction changing, but the rate of that change—or the acceleration—is increasing as time goes on. The changing acceleration indicates a changing velocity, which is key to understanding the behavior of the moving particle.

Instantaneous Power

The concept of "instantaneous power" involves how quickly work is done over time. In physics, power is the rate at which energy is transferred or converted. If you've ever ridden a bicycle up a hill, you know that you need to exert more effort (and thus more power) to keep going at the same speed.

In this exercise, the instantaneous power delivered to a particle moving in a circle is what we're interested in. It's calculated as the rate at which the particle's kinetic energy changes. Kinetic energy (KE) depends on the velocity of the particle, defined as \[ KE = \frac{1}{2} m v^2 \]To find the power \( P \), we differentiate KE with respect to time \( t \), which effectively shows us how KE is changing at any moment:\[ P = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) \]In this context, substituting the value of \( v^2 \) from our earlier calculation gives us the power delivered:\[ P = m k^2 r^2 t \]Thus, the instantaneous power helps us understand the rate of energy change that keeps the particle moving in its path.

Kinetic Energy

Kinetic energy is all about motion. When we talk about kinetic energy, we're really talking about the energy an object has because of its speed. Anything that's moving has kinetic energy—cars on a road, water flowing in a river, or a particle zipping around a circular path.

The kinetic energy \( KE \) for a particle of mass \( m \) moving at velocity \( v \) is given by the equation:\[ KE = \frac{1}{2} m v^2 \]In this exercise, since the velocity of the particle changes over time due to its centripetal acceleration, the kinetic energy also varies with time. Substituting the expression for velocity we identified earlier \( v = k r t \), the formula for kinetic energy becomes:\[ KE = \frac{1}{2} m (k^2 r^2 t^2) \]This expression shows how the kinetic energy increases as time \( t \) increases and highlights the direct relationship between motion (through velocity) and energy. The changing kinetic energy is central to calculating the power since it informs us about how energy is being transferred as the particle continues along its circular path. This dynamic makes kinetic energy crucial in understanding the behavior of moving objects.