A fly wheel of radius 10cm is mounted so as rotate about horizontal axis pass through it center. A string of negligible mass wrapped round it's circumference carrie's mass 0.5kg attached to it's free end .when the mass released from rest it fall through a distance of 1m in 2second. Find(a) the moment of inertia of the fly wheel about the axis of rotation
(b) the tension in string
To find the moment of inertia of the flywheel about the axis of rotation, we can use the principle of conservation of energy.
(a) The potential energy lost by the mass as it falls is converted into the rotational kinetic energy of the flywheel. We can calculate the potential energy lost by the mass using the formula:
Potential Energy Lost = m * g * h
Where:
m = mass of the object (0.5 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height through which the mass falls (1 m)
Therefore, Potential Energy Lost = 0.5 * 9.8 * 1 = 4.9 Joules
Since this potential energy is converted into rotational kinetic energy, we have:
Rotational Kinetic Energy = 1/2 * I * ω^2
Where:
I = moment of inertia of the flywheel
ω = angular velocity (which is related to the rotational speed)
Since the mass is released from rest, its initial angular velocity is 0. We can find the final angular velocity using the formula:
ω = Δθ / Δt
Where:
Δθ = angular displacement (assuming the string unwraps completely)
Δt = time taken for the mass to fall (2 seconds)
From the given conditions, the angular displacement (Δθ) is equal to the distance fallen by the mass divided by the radius of the flywheel:
Δθ = 1 m / 0.1 m = 10 radians
Plugging in the values into the equation for angular velocity:
ω = 10 radians / 2 seconds = 5 radians/s
Now, we can find the moment of inertia of the flywheel by rearranging the equation for rotational kinetic energy:
I = (2 * Rotational Kinetic Energy) / ω^2
I = (2 * 4.9) / (5)^2
I = 0.98 kg * m^2
Therefore, the moment of inertia of the flywheel about the axis of rotation is approximately 0.98 kg * m^2.
(b) To find the tension in the string, we can use Newton's second law for rotational motion:
Tension * radius = I * α
Where:
Tension = tension in the string (what we need to find)
radius = radius of the flywheel (0.1 m)
I = moment of inertia of the flywheel (0.98 kg * m^2)
α = angular acceleration
Since the initial angular velocity was 0 and the final angular velocity is 5 radians/s, the angular acceleration can be calculated using the formula:
α = (ω - ω0) / Δt
Where:
ω = final angular velocity (5 radians/s)
ω0 = initial angular velocity (0 radians/s)
Δt = time taken for the mass to fall (2 seconds)
Plugging in the values:
α = (5 radians/s - 0 radians/s) / 2 seconds
α = 2.5 radians/s^2
Now, we can find the tension in the string:
Tension * radius = I * α
Tension = (I * α) / radius
Tension = (0.98 * 2.5) / 0.1
Tension = 24.5 N
Therefore, the tension in the string is approximately 24.5 Newtons.