A uniform disc of radius 20 cm and mass 2 kg can rotate about a fixed axis through the centre and perpendicular to its plane. A massless cord is round along the rim of the disc. If a uniform force of2 newtons is applied on the cord, tangential acceleration of a point on the rim of the disc will be :-
Question
A uniform disc of radius 20 cm and mass 2 kg can rotate about a fixed axis through the centre and perpendicular to its plane. A massless cord is round along the rim of the disc. If a uniform force of2 newtons is applied on the cord, tangential acceleration of a point on the rim of the disc will be :-
Solution
The problem involves the concept of rotational motion. Here are the steps to solve it:
Step 1: Identify the given values.
- Radius (r) = 20 cm = 0.2 m (converted from cm to m)
- Mass (m) = 2 kg
- Force (F) = 2 N
Step 2: Calculate the moment of inertia (I) of the disc. The moment of inertia of a disc rotating about an axis perpendicular to its plane and passing through its center is given by the formula I = 0.5mr^2. Substituting the given values, we get:
I = 0.5 * 2 kg * (0.2 m)^2 = 0.02 kg*m^2
Step 3: Calculate the angular acceleration (α) using the formula for torque (τ). Torque is the rotational equivalent of force and is given by the formula τ = Iα. The torque due to the force applied on the cord is τ = Fr. Equating the two expressions for torque, we get:
Fr = Iα
Solving for α, we get:
α = Fr / I = 2 N * 0.2 m / 0.02 kgm^2 = 20 rad/s^2
Step 4: Calculate the tangential acceleration (a_t) of a point on the rim of the disc. The tangential acceleration is related to the angular acceleration by the formula a_t = r*α. Substituting the values, we get:
a_t = 0.2 m * 20 rad/s^2 = 4 m/s^2
So, the tangential acceleration of a point on the rim of the disc is 4 m/s^2.
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