Home
Class 12
PHYSICS
For a particle moving along a circular p...

For a particle moving along a circular path the radial acceleration `a_r ` is proportional to `t^2` (square of time) . If `a_z` is tangential acceleration which of the following is independent of time

A

`a_r . a_z`

B

`a_z`

C

`a_r//a_z`

D

`((a_r)^2)/(a_z)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the given information about the particle moving along a circular path. Here’s the step-by-step solution: ### Step 1: Understand the given information We know that the radial acceleration \( a_r \) is proportional to \( t^2 \). This can be expressed mathematically as: \[ a_r = c t^2 \] where \( c \) is a constant. ### Step 2: Relate radial acceleration to velocity In circular motion, the radial (or centripetal) acceleration is also given by the formula: \[ a_r = \frac{v^2}{r} \] where \( v \) is the linear velocity and \( r \) is the radius of the circular path. ### Step 3: Equate the two expressions for radial acceleration From the above two equations, we can equate: \[ c t^2 = \frac{v^2}{r} \] Rearranging gives us: \[ v^2 = c r t^2 \] ### Step 4: Find the expression for velocity Taking the square root of both sides, we find the expression for velocity: \[ v = \sqrt{c r} \cdot t \] This shows that the velocity \( v \) is directly proportional to time \( t \). ### Step 5: Determine the tangential acceleration The tangential acceleration \( a_z \) is defined as the rate of change of velocity with respect to time: \[ a_z = \frac{dv}{dt} \] Substituting the expression for \( v \): \[ a_z = \frac{d}{dt}(\sqrt{c r} \cdot t) = \sqrt{c r} \cdot \frac{d}{dt}(t) = \sqrt{c r} \] Since \( \sqrt{c r} \) is a constant (as both \( c \) and \( r \) are constants), we conclude that: \[ a_z = \sqrt{c r} \] is independent of time. ### Step 6: Analyze the options Now we need to check which of the given options is independent of time: 1. \( a_r \cdot a_z \): \( a_r \) is proportional to \( t^2 \) (depends on time), while \( a_z \) is constant. Thus, this product depends on time. 2. \( a_z \): As derived, \( a_z \) is constant and independent of time. This is a correct option. 3. \( \frac{a_r}{a_z} \): \( a_r \) depends on time, while \( a_z \) is constant. Thus, this ratio depends on time. 4. \( \frac{a_r^2}{a_z} \): Since \( a_r \) is proportional to \( t^2 \), \( a_r^2 \) is proportional to \( t^4 \). Therefore, this expression depends on time. ### Conclusion The only quantity that is independent of time is: \[ \boxed{a_z} \]

To solve the problem, we need to analyze the given information about the particle moving along a circular path. Here’s the step-by-step solution: ### Step 1: Understand the given information We know that the radial acceleration \( a_r \) is proportional to \( t^2 \). This can be expressed mathematically as: \[ a_r = c t^2 \] where \( c \) is a constant. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

For a particle moving along circular path, the radial acceleration a_x is proportional to time t. If a_t is the tangential acceleration, then which of the following will be independent of time t.

A particle is moving along a circular path of radius of R such that radial acceleration of particle is proportional to t^(2) then

A particle is moving on a circular path with constant speed, then its acceleration will be

A particle is moving along X-axis Its acceleration at time t is proportional to its velocity at that time. The differential equation of the motion of the particle is

If a particle moves such that the displacement is proportional to the square of the velocity acquired, then it acceleration is

The distance covered by a moving body is directly proportional to the square to the time. The acceleration of the body is

A point starts from rest and moves along a circular path with a constant tangential acceleration. After one rotation, the ratio of its radial acceleration to its tangential acceleration will be equal to