Animated Solution for Physics - Work, Energy, and Power: A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac=k2rt2, where k is a constant. The power delivered to the particle by the force acting on it is
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Visualized Solution
The Setup
Particle of mass m in a circle of radius r.
Centripetal acceleration is given as ac=k2rt2.
ac=rv2
We know that centripetal acceleration is related to speed v by the formula:
ac=rv2
rv2=k2rt2
Equating the given expression with the formula:
rv2=k2rt2
v=krt
Solving for v2:
v2=k2r2t2
Taking the square root gives the speed:
v=krt
at=dtdv
The particle's speed is changing with time, which means there is a tangential acceleration at.
at=dtdv
at=kr
Differentiating v=krt with respect to time t:
at=dtd(krt)
at=kr
Ft=mat
According to Newton's Second Law, the tangential force Ft is:
Ft=mat
Ft=mkr
Substituting the value of at:
Ft=m(kr)
Ft=mkr
P=F⋅v
Power delivered by a force is the dot product of force and velocity:
P=F⋅v
The centripetal force is perpendicular to velocity, so it does zero work.
P=Ftv
Only the tangential force is parallel to the velocity and delivers power:
P=Ftv
P=(mkr)(krt)
Substituting the expressions for Ft and v:
P=(mkr)(krt)
P=mk2r2t
Multiplying the terms together:
P=mk2r2t
This matches option (b).
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The Sigma Insight: Kinetic Energy, Potential Energy and Power
Solution Diagram
The beauty of circular motion lies not just in the fact that an object is going in circles, but in the intricate dance of forces that keep it there and dictate its speed. When a particle moves in a circular path, it experiences a fascinating interplay of kinematics and dynamics. Let's dive deep into this classic JEE problem and unravel the physics behind it.
The Setup
A Particle in a Circle
Imagine a particle of mass m constrained to move in a circular path of a constant radius r. We are given a very specific condition: its centripetal acceleration ac is not constant, but varies with time t according to the equation:
ac=k2rt2
Here, k is just a positive constant. What does this equation physically mean? Centripetal acceleration is the inward "pull" required to keep the particle turning. If this pull is increasing quadratically with time, it implies the particle must be spinning faster and faster. It's like driving a car in a circle and continuously pressing the accelerator pedal.
Unlocking the Velocity
To understand how fast the particle is moving, we need to connect the given centripetal acceleration to its speed. The fundamental kinematic relationship for circular motion tells us that centripetal acceleration is the square of the speed divided by the radius:
ac=rv2
By equating our theoretical formula with the given expression, we can unlock the velocity of the particle:
rv2=k2rt2
Multiplying both sides by r, we isolate v2:
v2=k2r2t2
Taking the positive square root (since speed is the magnitude of velocity and must be non-negative), we find:
v=krt
This is a crucial breakthrough! The speed of the particle is increasing linearly with time.
The Tangential Push
In circular motion, acceleration has two distinct jobs. The centripetal acceleration (ac) changes the direction of the velocity vector, keeping the particle on the circular track. However, because the speed v is changing, there must be another component of acceleration responsible for this change in magnitude. This is the tangential acceleration (at).
By definition, tangential acceleration is the rate of change of speed:
at=dtdv
Substituting our expression for v:
at=dtd(krt)=kr
Since k and r are constants, the tangential acceleration is constant! Now, according to Newton's Second Law, a tangential acceleration requires a tangential force (Ft).
Ft=mat=mkr
So, a constant tangential force is continuously pushing the particle along the circular track, causing its speed to increase linearly.
The Power Play
The ultimate goal of the problem is to find the power delivered to the particle. Power is defined as the rate at which work is done, which can be elegantly expressed as the dot product of the net force vector and the velocity vector:
P=Fnet⋅v
The net force on the particle has two components: the centripetal force (Fc) and the tangential force (Ft).
Here is the catch: The centripetal force always points radially inward, exactly perpendicular (90∘) to the instantaneous velocity vector. Because the dot product of perpendicular vectors is zero (cos90∘=0), the centripetal force does zero work and delivers zero power. It merely steers the particle.
Therefore, all the power is delivered exclusively by the tangential force, which is perfectly parallel to the velocity:
P=Ftv
Substituting the values we derived earlier:
P=(mkr)(krt)
P=mk2r2t
The Ninja Technique
Work-Energy Theorem
While the force method is highly instructive, there is a much faster, more elegant way to solve this using the Work-Energy Theorem. The theorem states that the net work done on a particle equals its change in kinetic energy. Consequently, Power (the rate of doing work) is simply the rate of change of kinetic energy!
First, write down the kinetic energy (K) using the speed we found (v=krt):
K=21mv2=21m(krt)2=21mk2r2t2
Now, just differentiate this kinetic energy with respect to time to get the power:
P=dtdK=dtd(21mk2r2t2)
P=21mk2r2(2t)=mk2r2t
In just two lines of calculus, we arrive at the exact same result! This highlights the profound consistency and beauty of physics. Whether you analyze the forces or track the energy, the universe always balances its books.