Question

A puck of mass $\mathbf{m}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{kg}$slides in a circle of radius$\mathbf{r}\mathbf{=}\mathbf{}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{cm}$on a frictionless table while attached to a hanging cylinder of mass$\mathbf{M}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{kg}$by means of a cord that extends through a hole in the table (Fig. 6-43). What speed keeps the cylinder at rest?

Short answer

The speed that keeps the cylinder at rest is $1.81\text{\hspace{0.17em}m/s}$.

Step-by-step solution

Given information

• Mass of pluck,$\mathrm{m}=1.50\text{kg}$.
• Mass of hanging cylinder,$\mathrm{M}=2.50\text{kg}$.
• The radius of the circle, $\mathrm{r}=20.0\text{cm}$.

To understand the concept

The problem deals with the centripetal force. It is a force that makes a body follow a curved path.For the puck to remain at rest, the magnitude of the tension force$\mathbf{T}$of the cord must equal the gravitational force$\mathbf{Mg}$on the cylinder. The tension force supplies the centripetal force that keeps the puck in its circular orbit.

To calculate the speed that keeps the cylinder at rest

In the equilibirum condition:

$\begin{array}{c}\text{Tensionforce}=\text{Centripetalforce}\\ \mathrm{T}={\mathrm{mv}}^2/\mathrm{r}\end{array}$

But,

$\mathrm{T}=\mathrm{Mg}$

Thus,

$\mathrm{Mg}={\mathrm{mv}}^2/\mathrm{r}$

We solve for the speed as:

$\mathrm{v}=\sqrt{\frac{\mathrm{Mgr}}{\mathrm{m}}}$

Substitute the values, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{\frac{\left(2.50\text{\hspace{0.17em}kg}\right)\left(9.80{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\left(0.200\text{\hspace{0.17em}m}\right)}{1.50\text{\hspace{0.17em}kg}}}\\ \mathrm{v}=1.81\text{\hspace{0.17em}m/s}\end{array}$

Thus, the speed that keeps the cylinder at rest is $1.81\text{\hspace{0.17em}m/s}$