Question

A 200 g ball moves in a vertical circle on the end of a 50 cm long string. If its speed at the bottom is 10 m/s, calculate: (a) the velocity at the top of the circle. (b) the tension at the top of the circle.

Answer 1

(a) The velocity at the top of the circle is approximately
`\(6.07 \, \text{m/s}\)`.

(b) The tension at the top of the circle is approximately
`\(16.3 \, \text{N}\)`.

In a vertical circular motion, the speed of the object changes as it moves around the circle. At the bottom of the circle, the tension in the string provides the centripetal force required for circular motion, and at the top, gravity provides the centripetal force. Let's solve for the velocity at the top and the tension at the top.

Given:

- Mass of the ball
`(\(m\))`= 200 g = 0.2 kg

- Length of the string
`(\(r\))` = 50 cm = 0.5 m

- Speed at the bottom
`(\(v_{\text{bottom}}\))` = 10 m/s

First, let's find the velocity at the top
`(\(v_{\text{top}}\))`. The conservation of energy can be applied, considering the conversion of potential energy to kinetic energy:

`\[ mgh_{\text{bottom}} + (1)/(2)mv_{\text{bottom}}^2 = mgh_{\text{top}} + (1)/(2)mv_{\text{top}}^2 \]`

At the bottom of the circle,
`\(h_{\text{bottom}} = 0\)` (since the reference point is usually at the bottom), and at the top of the circle,
`\(h_{\text{top}} = 2r\)`.

`\[ (1)/(2)mv_{\text{bottom}}^2 = mg(2r) + (1)/(2)mv_{\text{top}}^2 \]`

Now, solve for
`\(v_{\text{top}}\)`.

`\[ v_{\text{top}} = \sqrt{v_{\text{bottom}}^2 - 2gr} \]`

Substitute the known values:

`\[ v_{\text{top}} = \sqrt{(10 \, \text{m/s})^2 - 2 * 9.8 \, \text{m/s}^2 * 0.5 \, \text{m}} \]`

`\[ v_{\text{top}} \approx 6.07 \, \text{m/s} \]`

Next, let's find the tension at the top
`(\(T_{\text{top}}\))`. At the top of the circle, the net force acting on the ball is the centripetal force provided by tension, so:

`\[ T_{\text{top}} = \frac{mv_{\text{top}}^2}{r} + mg \]`

`\[ T_{\text{top}} = \frac{0.2 \, \text{kg} * (6.07 \, \text{m/s})^2}{0.5 \, \text{m}} + 0.2 \, \text{kg} * 9.8 \, \text{m/s}^2 \]`

`\[ T_{\text{top}} \approx 16.3 \, \text{N} \]`

Answer 2

(a) The velocity at the top of the circle is 10.5 m/s.

(b) The tension at the bottom of the circle is 41.96 N.

How to calculate the velocity?

(a) The velocity at the top of the circle is calculated by applying the following formula.

T + mg = mv²/r

where;

• T is the tension at the top of the circle
• m is the mass of the ball
• g is gravity
• v is the speed of the ball
• r is the radius of the path

Since we have the speed at the bottom of the circle, the tension is calculated as;

Equation for net force at the bottom is;

T - mg = mv²/r

T = mv²/r + mg

T = (0.2 kg x (10 m/s)² ) / 0.5 m + (0.2 kg x 9.8 m/s²)

T = 41.96 N

T + mg = mv²/r

41.96 + (0.2 x 9.8) = (0.2v²)/0.5

43.92 = 0.4v²

v² = 109.8

v = √(109.8)

v = 10.5 m/s

(b) The tension at the bottom of the circle is already calculated as 41.96 N.