Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Circular Motion
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Circular Motion
A hemispherical bowl of radius R is rotated about its vertically...
A hemispherical bowl of radius is rotated about its vertically maintained axis of symmetry. A small block is placed inside the bowl at a position where the radius vector forms an angle with the vertical. The block rotates alongside the bowl without experiencing any slipping. The coefficient of friction between the surface of the bowl and the block is . Determine the range of the angular speed for which the block avoids slipping.
Options
- A to
- B to
- C to
- D to
Correct answer
A. to
Step-by-step solution
Consider a block of mass rotating in a horizontal circle of radius .
The forces acting on the block are the normal reaction , the weight , and the frictional force .
For the minimum angular speed \omega_\min, the block has a tendency to slip downwards, so the limiting frictional force acts upwards along the tangent to the bowl.
Balancing forces in the vertical direction:
The net force towards the center provides the centripetal acceleration:
N\sin\theta - f\cos\theta = m\omega_\min^2 r
Using r = R\sin\thetaN(\cos\theta + \mu\sin\theta) = mgN(\sin\theta - \mu\cos\theta) = m\omega_\min^2 R\sin\theta\dfrac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta} = \dfrac{\omega_\min^2 R\sin\theta}{g}\omega_\min = \left[ \dfrac{g(\sin\theta - \mu\cos\theta)}{R\sin\theta(\cos\theta + \mu\sin\theta)} \right]^{1/2}\omega_\maxN\cos\theta - f\sin\theta = mgN\sin\theta + f\cos\theta = m\omega_\max^2 rf = \mu Nr = R\sin\thetaN(\cos\theta - \mu\sin\theta) = mgN(\sin\theta + \mu\cos\theta) = m\omega_\max^2 R\sin\theta\dfrac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} = \dfrac{\omega_\max^2 R\sin\theta}{g}\omega_\max =
The forces acting on the block are the normal reaction , the weight , and the frictional force .
For the minimum angular speed \omega_\min, the block has a tendency to slip downwards, so the limiting frictional force acts upwards along the tangent to the bowl.
Balancing forces in the vertical direction:
The net force towards the center provides the centripetal acceleration:
N\sin\theta - f\cos\theta = m\omega_\min^2 r
Using r = R\sin\thetaN(\cos\theta + \mu\sin\theta) = mgN(\sin\theta - \mu\cos\theta) = m\omega_\min^2 R\sin\theta\dfrac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta} = \dfrac{\omega_\min^2 R\sin\theta}{g}\omega_\min = \left[ \dfrac{g(\sin\theta - \mu\cos\theta)}{R\sin\theta(\cos\theta + \mu\sin\theta)} \right]^{1/2}\omega_\maxN\cos\theta - f\sin\theta = mgN\sin\theta + f\cos\theta = m\omega_\max^2 rf = \mu Nr = R\sin\thetaN(\cos\theta - \mu\sin\theta) = mgN(\sin\theta + \mu\cos\theta) = m\omega_\max^2 R\sin\theta\dfrac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} = \dfrac{\omega_\max^2 R\sin\theta}{g}\omega_\max =