A50 gram yo-yo is released and travels vertically downward, rolling without slipping. knowing that the yo-yo’s moment of inertia is 9 x 10-5 kg m2 and the radius of its center spindle is 0.7 cm,
find
a) the torque involved and
b) the angular acceleration of the yo-yo.

Explanation:

mass, m = 50 g

moment of inertia, I = 9 x 106-5 kg m^2

radius, r = 0.7 cm

(a) As it moving downwards

Let T be the tension in the string

T = m (g + a)   (1)

where, a be the acceleration

τ = I α = T r

α =  a / r

So, I x a / r = T x r

a = T r^2 / I

Substitute in equation (1) we get

a = m (g + a) r^2 / I

a = mgr^2 / (I - mr^2)

a = 0.050 x 9.8 x 0.007 x 0.007 / (9 x 10^-5 - 0.050 x 0.007 x 0.007)

a = 2.401 x 10^-5 / (87.55 x 10^-6)

a = 0.274 m/s^2

τ = I x α = I x a / r

τ = 9 x 10^-5 x 0.274 / 0.007

τ = 3.52 x 10^-3 Nm

(b) α = a / r

α = 0.274 / 0.007 = 39.14 rad/s^2

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