Question

A ring of mass \(\mathrm{m}\) and radius \(\mathrm{R}\) is rolling without sliding on a flat fixed horizontal surface, with a constant angular velocity. The angle between the velocity and acceleration vectors of a point \(\mathrm{P}\) on the rim of the ring at the same horizontal level as the centre of the ring may be - (1) zero (2) \(90^{\circ}\) (3) \(135^{\circ}\) (4) \(\tan ^{-1} 1 / 2\)

Short answer

90 degrees (2)

Step-by-step solution

- Analyze Motion of Point P

Consider point \(P\) on the rim of the ring, which is at the same horizontal level as the center of the ring. Since the ring is rolling without slipping, point \(P\) has both translational and rotational motion.

- Determine Velocity of Point P

The translational velocity \(v_t\) of the ring's center is \(v = \omega R = \text{constant}\). For point \(P\) on the rim, the velocity due to rotation is also \(\omega R\). Therefore, the velocity vector \(\vec{v_P}\) of point \(P\) at the moment it is at the same height as the center is directed tangentially and horizontally forward with a magnitude of \(2\omega R\).

- Calculate Acceleration of Point P

Point \(P\) experiences two accelerations: centripetal acceleration \(a_c = \omega^2 R\) directed towards the center of the ring, and tangential acceleration \(a_t = 0\) because angular velocity \(\omega\) is constant.

- Determine Direction of Acceleration Vectors

The centripetal acceleration vector is directed towards the center of the ring, which is perpendicular to the direction of the velocity vector for point \(P\) when it is at the horizontal level.

- Calculate Angle Between Velocity and Acceleration

Since the tangential acceleration \(a_t = 0\) and the only acceleration acting is the centripetal acceleration \(a_c\), the angle \(\theta\) between the velocity and acceleration vectors of point \(P\) is \(90^{\circ}\).

Key concepts

Centripetal Acceleration

When an object follows a curved path or circular motion, it constantly changes direction even if its speed remains constant. This change in direction results in centripetal acceleration, which always points towards the center of the circle. For a ring rolling without slipping, every point on the rim will experience this inward acceleration.
Centripetal acceleration is calculated using the formula: \[ a_c = \frac{v^2}{R} = \text{constant} \]
or equivalently \[ a_c = \text{constant} = \text{\textsl{constant}} = \text{\texttheta} \]
In this exercise, the point P on the rim of the ring at the same height as the center will experience centripetal acceleration directed towards the center. This is the only acceleration component since tangential acceleration is zero. This perpendicular direction to the velocity results in a 90-degree angle between the velocity and acceleration vectors.

Rotational Dynamics

Rotational dynamics involves the study of objects in rotational motion and the forces that cause or result from this motion. For a rolling ring, the key aspect is that it rolls without slipping, combining translational and rotational motion.
Key Points in Rotational Dynamics:

• Angular velocity (ω): How fast the object rotates. It is constant in this problem.
• Translational velocity (v): The speed of the center of mass, given by \( v = \text{\textsl{constant}} \).
• Moment of Inertia (I): Reflects how mass is distributed relative to the axis of rotation. For a ring, \( I = mR^2 \).

The total velocity of point P on the rim is a combination of translational and rotational velocities, resulting in \( v_P = 2ωR \). This understanding helps in identifying the instantaneous velocity and acceleration vectors.

Angular Velocity

Angular velocity is a measure of how fast an object rotates around an axis. It is denoted by \(ω\) and for a rolling ring, it determines both the rotational speed and its translational motion.
Important aspects of Angular Velocity:

• It is a vector quantity, having both magnitude and direction.
• For rolling without slipping, the relation between angular velocity and translational velocity is \( v = ωR \).
• If angular velocity is constant, it implies no tangential acceleration.

In the given exercise, the ring’s angular velocity remains constant, thus simplifying the problem. This implies that the tangential acceleration component is zero, leaving only centripetal acceleration to consider, which leads us to the 90-degree angle between the velocity and acceleration vectors of point P.