Answer :
Answer:
Acceleration of the center of the disc:
a = (T - Mg + N) / M
Angular acceleration of the disc:
α = 2 × T / (M × r)
Since N = Mg (equilibrium) and g = 9.8 m/s^2 (standard gravity), we can simplify the expressions:
a = (T - Mg + Mg) / M = T / M
α = 2 × T / (M × r)
So, the answers are:
a = T / M
α = 2 × T / (M × r)
Please note that these are general expressions, and you'll need to plug in specific values for T, M, and r to get numerical answers.
Explanation:
classic problem in mechanics!
Let's break it down step by step:
1. ** Forces acting on the disc **:
- Force of gravity (Mg) acting downwards
- Normal force (N) acting upwards from the rough surface
- Force of tension (T) in the thread, acting tangentially
- Friction force (f) acting opposite to the motion
2. ** Motion of the disc **:
- The disc rolls without slipping, so the point of contact with the surface is at rest.
- The center of the disc moves with acceleration (a)
- The disc rotates with angular acceleration (α)
3. ** Equations of motion **:
- For linear motion: Mg - N + T = Ma (since f = 0, no slipping)
For rotational motion: T × r = I × α (where I is the moment of inertia of the disc)
4. ** Moment of inertia **:
- For a uniform disc: I = (1/2) × M × r^2
5. ** Solving for a and α **:
- From the linear motion equation: a = (T - Mg + N) / M
- From the rotational motion equation: α = T × r / I = T × r / (0.5 × M × r^2) = 2 × T / (M × r)
So, the acceleration of the center of the disc is:
a = (T - Mg + N) / M
And the angular acceleration of the disc is:
α = 2 × T / (M × r)
Note that N = Mg if the disc is in equilibrium, and f = 0 since it rolls without slipping. You can plug in the values of T, M, r, and g to find the numerical values of a and α.
