Question
A small block B of mass m = 0.25 kg, lying on a frictionless horizontal table, is attached to a massless cord (breaking strength 40 N) passing through a narrow hole C at the centre of the table. Initially when the block is revolving in a circle of radius r0 = 0.80 m about a vertical axis through the hole, with a tangential speed of v0 = 4.00 m/s; the tension in the string is T0 and the kinetic energy of the block is K0. The string is then pulled down slowly from below, decreasing the radius of circular path from r0 to r so that the kinetic energy of the block is now K and the tension in the string is T. As a result :-
Options
Solution
[Key Concept: A central (radial) pull exerts no torque, so the block's angular momentum about the hole is conserved.]
(a) angular momentum conservation
$$\begin{matrix} & {mv}_{0}r_{0} = mvr \\ & v = \frac{v_{0}r_{0}}{r} \\ & {\text{ }T}_{0} = \frac{{mv}_{0}^{2}}{r_{0}} \\ & \text{ }T = \frac{{mv}^{2}}{r} = \frac{{mv}_{0}^{2}r_{0}^{2}}{r^{3}} \\ & \ \Rightarrow \text{ }T = \frac{T_{0}r_{0}^{3}}{r^{3}} \end{matrix}$$
b) $\ k = \frac{1}{2}{mv}^{2} = \frac{1}{2}\text{ }m\frac{v_{0}^{2}r_{0}^{2}}{r^{2}} = \frac{k_{0}r_{0}^{2}}{r^{2}}$
c)
$$\begin{matrix} & \text{~~}T = \frac{T_{0}r_{0}^{3}}{r^{3}} = 40 \\ & \ \Rightarrow \frac{{mv}_{0}^{2}r_{0}^{2}}{r^{3}} = 40 \\ & \ \Rightarrow \frac{0.25 \times 16 \times 0.64}{40} = r^{3} \\ & \ \Rightarrow r = 0.4\text{ }m \end{matrix}$$
d)
$$\begin{matrix} & \text{~}W = \Delta K = k_{f} - k_{i} \\ & \ = \frac{k_{0}r_{0}^{2}}{\left( \frac{r_{0}}{2} \right)^{2}} - k_{0} = 3k_{0} \end{matrix}$$
Create a free account to view solution
View Solution Free