A wheel of radius 6cm is mounted so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped round its circumference carries a mass of 200g attached to its free end. When allowed to fall, the mass descends through 100cm in 5 seconds. Calculate the angular acceleration of the wheel, its moment of inertia and the tension in the cord.
answer
0.1176
To solve this problem, we need to apply the principles of rotational motion. Let's break it down step by step:
Step 1: Find the linear acceleration of the mass
The given information tells us that the mass descends through a vertical distance of 100 cm in 5 seconds. Using the equation of motion, s = ut + (1/2)at^2, where s is the distance, u is initial velocity (zero in this case), t is time, and a is acceleration, we can solve for the linear acceleration. Rearranging the equation, we have:
100 = (1/2)at^2
Simplifying,
a = (2 * 100) / (5^2)
= 8 m/s^2 (since 1 cm = 0.01 m and 1 s = 1 m/s)
Step 2: Determine the angular acceleration
The linear acceleration of the mass is related to the angular acceleration of the wheel. They are connected by the equation a = rα, where a is the linear acceleration, r is the radius of the wheel, and α is the angular acceleration. Substituting the given values, we have:
8 = (0.06)α (since the radius of the wheel is given in centimeters)
α = 8 / 0.06
α ≈ 133.33 rad/s^2
Step 3: Calculate the moment of inertia
The moment of inertia (I) represents the object's resistance to changes in rotation. For a solid disk rotating about its central axis, the moment of inertia is given by I = (1/2)MR^2, where M is the mass of the wheel and R is its radius. Since the radius is given in centimeters, we need to convert it to meters to maintain consistency in units:
I = (1/2) * (0.2 kg) * (0.06 m)^2
I ≈ 0.00036 kg·m^2
Step 4: Determine the tension in the cord
The tension in the cord can be found using the equation τ = Iα, where τ is the torque applied to the wheel. In this case, the torque is equal to the tension in the cord multiplied by the radius of the wheel. Thus,
τ = T * r
Using τ = Iα, we can solve for T:
T * 0.06 = 0.00036 * 133.33
T ≈ 2.401 N (approximately)
So, the angular acceleration of the wheel is approximately 133.33 rad/s^2, its moment of inertia is approximately 0.00036 kg·m^2, and the tension in the cord is approximately 2.401 N.