Chapter 3: Problem 47
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
Step by step solution
Identify the Problem
Understand Rolling Without Slipping
Describe Velocity of Point P
Calculate Total Velocity of Point P
Determine The Acceleration Components
Find Direction of Acceleration Vector
Calculate Angle Between Velocity and Acceleration
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
angular velocity
To understand angular velocity better:
- Imagine a point on the rim of the ring moving in a circular path due to rotation.
- The faster the ring spins, the greater the angular velocity.
- Angular velocity is crucial for calculating other dynamic properties like rotational kinetic energy and centripetal acceleration.
velocity vectors
- Translational velocity: The speed at which the center of the ring moves horizontally, given by \(\text{Rω}\).
- Rotational velocity: The speed due to the ring’s own rotation, also given by \(\text{Rω}\).
\(\text{v}_{total} = 2 \text{R} \text{ω}\).
Velocity vectors help visualize how an object moves. Directionality is key:
- A vector pointing forward indicates forward motion.
- Adding vectors ensures a realistic representation of combined movements.
centripetal acceleration
In the exercise, point P only experiences centripetal acceleration since the ring’s angular velocity (\text{ω}) is constant and there’s no tangential (linear) acceleration.
Centripetal acceleration characteristics include:
- Always directed inward, towards the center of rotation.
- Magnitude proportional to the square of angular velocity and radius ( \(\text{R} \text{ω}^2\).