Question

AuniformsphericalshellofmassM=4.5kgandradiusR=8.5cmcan rotate about a fixed vertical axis on frictionless bearings. A massless, stretchless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.0 x 10‐3 kg m2 and radius r = 5.0 cm, and is attached to a small object of mass m = 0.60 kg, which hangs on the end of the cord. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen 82 cm after being released from rest? Use energy considerations.

Answer 1

`v=1.42(m)/(s)`

Step-by-step explanation:

There is no friction in the physical system. Thus, according to the law of conservation of energy, recall that the object is released from rest:

`\Delta E=0\\U=K_R+K_T\\mgh=(I_p\omega_p^2)/(2)+(I_s\omega_s^2)/(2)+(mv^2)/(2)`

Recall that the moment of inertia of a sphere is
`I_s=(2MR^2)/(3)`. The angular speed of the pulley is
`\omega_p=(v)/(r)` and the angular speed of the sphere is
`\omega_s=(v)/(R)`. So, we replace:

`mgh=(I_pv^2)/(2r^2)+(2MR^2)/(3)(v^2)/(2R^2)+(mv^2)/(2)\\mgh=v^2((I_p)/(2r^2)+(M)/(3)+(m)/(2))\\v^2=(mgh)/((I_p)/(2r^2)+(M)/(3)+(m)/(2))\\\\v=\sqrt{((0.6kg)(9.8(m)/(s^2))(0.82m))/((3*10^(-3)kg\cdot m^2)/(2(0.05m)^2)+(4.5kg)/(3)+(0.6kg)/(2))}\\v=1.42(m)/(s)`