Question

A 2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is θ = 30­^o.
a. What is the ball’s speed when θ = 30^o­?
b. What are the magnitude and direction of the ball’s acceleration when θ = 30^o­?

Short answer

a) Balls speed is 3.85 m /sec

b) Ball's acceleration is 18.496 m/sec²

Step-by-step solution

Part(a) Step 1 : Given Information

Mass of ball is 2 kg
Length of string = 80 cm =0.8 m
Tension in the string = 20 N
When its angle from the highest point on the circle is θ = 30^o­.

Part(a) Step 2: Explanation

The centripetal force acting on a mass m revolving with speed v around a circle of radius r is given by

mv²/r ……………………………….(1)

The centripetal acceleration is given by

v²/r ……………………………….(2)

The tension of the spring at a given angle is and solve for v

$$\begin{gathered} T=\frac{m{v}^2}{r}-mgcos(30°) \\ \frac{m{v}^2}{r}=T+mgcos(30°) \\ {v}^2=\frac{r(T+mg\cos (30°\left)\right)}{m} \\ v=\sqrt{\frac{r(T+mg\cos (30°\left)\right)}{m}} \end{gathered}$$

Substitute the given value we get

$v=\sqrt{\frac{(0.8m)\left(20N\right)+\left(2kg\right)(9.8m/s^2)\cos (30°)}{2kg}}=3.85m/s$

Part(b) Step 1: Given information

Mass of ball is 2 kg
Length of string = 80 cm =0.8 m
Tension in the string = 20 N
When its angle from the highest point on the circle is θ = 30^o­.

Part(b) Step 2: Explanation

The ball's radial acceleration is calculated as

$a=\frac{v^2}{r}$

Substitute the values given

$a=\frac{{(3.85m/s)}^2}{(0.8m)}=18.496m/s^2$