user

A ball of mass 0.35 kg is attached to the end of a horizontal cord and is rotated in a circle
of radius 1.0 m on a frictionless horizontal surface. If the cord breaks when the tension in
it exceeds 80 N, determine:
(a) the maximum speed of the ball,
(b) the minimum period of the ball.

alphonsio

(a) The maximum speed of the ball is approximately $15.11 \, \text{m/s}$.

(b) The minimum period of the ball is approximately $0.416 \, \text{s}$.

To solve this problem, we’ll first use the relationship between centripetal force, mass, velocity, and radius. The centripetal force $F_c$ required to keep an object of mass $m$ moving in a circle of radius $r$ with a speed $v$ is given by:

$F_c = \frac{mv^2}{r}$

Given:

- Mass, $m = 0.35 \, \text{kg}$
- Radius, $r = 1.0 \, \text{m}$
- Maximum tension (which is the centripetal force at breaking point), $F_c = 80 \, \text{N}$

We need to determine the maximum speed $v_{\text{max}}$ at which the cord can hold the ball without breaking. Using the centripetal force formula:

$v^2 = \frac{r \times F_c}{m}$

$v = \sqrt{ \frac{r \times F_c}{m}}$

Solving for $v_{\text{max}}$:

$v_{max} = \sqrt{ \frac{1 \, \text{m} \times 80 \, \text{N}}{0.35 \, \text{kg}}}$

$v_{\text{max}} \approx \sqrt{228.57}$

$v_{\text{max}} \approx 15.1185789204 \, \text{m/s}$

The period $T$ is the time it takes for the ball to make one complete revolution around the circle. The relationship between the speed $v$, the period $T$, and the radius $r$ of the circle is given by:

$v = \frac{2\pi r}{T}$

Solving for the period $T$:

$T = \frac{2\pi r}{v}$

Using the maximum speed $v_{\text{max}}$:

$T_{\text{min}} = \frac{2\pi \cdot 1.0 \, \text{m}}{15.11 \, \text{m/s}}$

$T_{\text{min}} \approx \frac{2\pi}{15.11}$

$T_{\text{min}} \approx \frac{6.28}{15.11}$

$T_{\text{min}} \approx 0.4155936441 \, \text{s}$