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

Answer:

$v=1.42\frac{m}{s}$

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:

$\Delta E=0\\U=K_R+K_T\\mgh=\frac{I_p\omega_p^2}{2}+\frac{I_s\omega_s^2}{2}+\frac{mv^2}{2}$

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

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