A disk with a mass of 1300 g and radius 0.7 m has a block of mass 170 g hanging from it, as shown. The wheel and block start at rest. The moment that the block is released, what is the magnitude of the angular acceleration of the disk? (Note: the moment of inertia of a disk about its axis is 0.5MR2.)
To find the magnitude of the angular acceleration of the disk when the block is released, we can use the principle of torque and moment of inertia.
1. First, let's find the torque due to the hanging block. The torque is given by the product of the force applied and the lever arm distance, which is the radius of the disk. In this case, the force applied is the weight of the block, given by its mass multiplied by the acceleration due to gravity (9.8 m/s^2). So, the torque is T = (0.170 kg + 1.300 kg) × 9.8 m/s^2 × 0.7 m.
2. The moment of inertia of a disk about its axis is given by the formula I = 0.5MR^2, where M is the mass of the disk and R is the radius of the disk. Substituting the given values, we have I = 0.5 × 1.300 kg × (0.7 m)^2.
3. Now, we can use Newton's second law for rotational motion, which states that the torque is equal to the moment of inertia multiplied by the angular acceleration (T = Iα). Rearranging this equation, we get α = T / I.
4. Substituting the values from step 1 and step 2, we have α = (0.170 kg + 1.300 kg) × 9.8 m/s^2 × 0.7 m / (0.5 × 1.300 kg × (0.7 m)^2).
5. Simplifying the equation, we find α = 0.095 rad/s^2.
Therefore, the magnitude of the angular acceleration of the disk when the block is released is 0.095 rad/s^2.