A solid cylinder of mass 2 kg and radius 0.5 m is initially at rest. A constant

A solid cylinder of mass 2 kg and radius 0.5 m is initially at rest. A constant force of 10 N is applied tangentially to the edge of the cylinder for 5 seconds. Calculate the angular velocity of the cylinder after 5 seconds, assuming no slipping occurs.”
Solution:
1. First, let’s calculate the torque applied to the cylinder using the formula:
[ tau = r times F ]
Where:
– ( r ) is the radius of the cylinder (0.5 m)
– ( F ) is the applied force (10 N)
So,
[ tau = (0.5 , text{m}) times (10 , text{N}) = 5 , text{Nm} ]
2. Now, let’s calculate the moment of inertia of the cylinder using the formula for a solid cylinder:
[ I = frac{1}{2} m r^2 ]
Where:
– ( m ) is the mass of the cylinder (2 kg)
– ( r ) is the radius of the cylinder (0.5 m)
So,
[ I = frac{1}{2} (2 , text{kg}) (0.5 , text{m})^2 = 0.25 , text{kg} cdot text{m}^2 ]
3. Now, we can use Newton’s second law for rotation to relate torque, moment of inertia, and angular acceleration:
[ tau = I alpha ]
Where:
– ( alpha ) is the angular acceleration
Rearranging for ( alpha ), we get:
[ alpha = frac{tau}{I} ]
[ alpha = frac{5 , text{Nm}}{0.25 , text{kg} cdot text{m}^2} = 20 , text{rad/s}^2 ]
4. Now, we can use the angular acceleration to find the angular velocity after 5 seconds using the kinematic equation:
[ omega_f = omega_i + alpha t ]
Since the cylinder is initially at rest, ( omega_i = 0 ). So,
[ omega_f = 0 + (20 , text{rad/s}^2) times 5 , text{s} = 100 , text{rad/s} ]
Therefore, the angular velocity of the cylinder after 5 seconds is 100 rad/s.