a wheel radius 0.25 m is mounted on a frictionless horizontal axle. the moment of inertia of the wheel about the axis is 0.040. a light cord wrapped around the wheel supports the 0.50 kg of box. the wheel is released from the rest. what is the magnitude of acceleration of the 0.50 kg box
To find the magnitude of acceleration of the 0.50 kg box, we can use the concept of torque and rotational dynamics.
Step 1: Calculate the torque caused by the gravitational force acting on the box.
- The gravitational force acting on the box is given by F_gravity = mg, where m is the mass of the box (0.50 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The torque caused by this gravitational force can be calculated as τ = r * F_gravity, where r is the radius of the wheel (0.25 m).
τ = (0.25 m) * (0.50 kg) * (9.8 m/s^2)
= 1.225 N·m
Step 2: Use the torque to calculate the angular acceleration of the wheel.
- The torque can also be calculated as τ = I * α, where I is the moment of inertia of the wheel and α is the angular acceleration.
- Rearranging the equation, we get α = τ / I.
α = (1.225 N·m) / (0.040 kg·m^2)
= 30.625 rad/s^2
Step 3: Convert the angular acceleration to linear acceleration.
- The linear acceleration of a point on the rim of the wheel can be related to the angular acceleration by the equation a = r * α, where a is the linear acceleration and r is the radius of the wheel.
a = (0.25 m) * (30.625 rad/s^2)
= 7.65625 m/s^2
So, the magnitude of acceleration of the 0.50 kg box is approximately 7.66 m/s^2.