Question

A light string is wrapped around the outside of a 2.0-kg-wheel whose radius is 75 cm. The wheel has a frictionless axel that allows it to rotate but prevents its center of mass from moving. Assume the moment of inertia of the wheel is the same as that of a point particle of equal mass at the same radius from the axel. The string is then attached to a 3.0-kg hanging mass that is released from rest. While the mass is falling, what is the (a) angular acceleration of the wheel and (b) the tension in the string? (c) After the mass has fallen 1.5 m, what is the translational acceleration of a point on the rim of the wheel, in terms of a radial and tangential component?

Answer 1

Alpha = a/R I = Mr^2 = 2 times (0.75)^2 kg m^2 = 1.125 kg - m^2 Free body diagram of block Force equation on block mg - T = ma (1) Force equation on wheel tau = I alpha Tr = I a/r Tr^2 = Ia (2) from (1) & (2) eliminating T mg - Ia/r^2 = ma = > a (I/r^2 + m) = mg a = mg/(I/r^2 + m) and T = IA/r^2 = Img/r^2 (I/r^2 + M) T = mgI/I + mr^2 \r\n Thus alpha = a/r = mg/r (I/r^2 + m) = mg/I/r + mr = 3 times 9.8/1.125/.75 + 3 times .75 TildeEqual 7.84 rad/s^2 Tr = I alpha = > T = I alpha/r T = 1.125 times 7.84/.75 TildeEqual 11.76 N For wheel a'_r = v^2/r towards center a'_t = alpha r Now to find v, let's apply equation motion on block y = 1.5, u = 0