Question

To make a pendulum, a 210 g ball is attached to one end of a string that has a length of 1.2 m and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of 24° with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of 12° with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

Answer 1

a)
`v \approx 1.233\,(m)/(s)`, b)
`v \approx 1.426\,(m)/(s)`, c)
`\theta_(2) \approx 22.61^(\textdegree)`

Step-by-step explanation:

a) The speed of the ball is determined by applying the Principle of Energy Conservation:

`U_(g,1) + K_(1) = U_(g,2) + K_(2)`

The speed of the ball when the string makes an angle of 12° with the vertical is:

`K_(2) = U_(g,1) - U_(g,2) + K_(1)`

`(1)/(2)\cdot m \cdot v^(2) = m\cdot g \cdot L \cdot [(1-\cos \theta_(1))-(1 -\cos \theta_(2))]`

`v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_(2) - \cos \theta_(1))}`

`v = \sqrt{2\cdot (9.807\,(m)/(s^(2)) )\cdot (1.2\,m)\cdot (\cos 12^(\textdegree)-\cos 24^(\textdegree))}`

`v \approx 1.233\,(m)/(s)`

b) The maximum speed of the ball is:

`v = \sqrt{2\cdot g \cdot L \cdot (\cos \theta_(2) - \cos \theta_(1))}`

`v = \sqrt{2\cdot (9.807\,(m)/(s^(2)) )\cdot (1.2\,m)\cdot (\cos 0^(\textdegree)-\cos 24^(\textdegree))}`

`v \approx 1.426\,(m)/(s)`

c) The angle between the string and the vertical when the speed of the ball is one-third its maximum value is obtained by proving different values of
`\theta_(2)`. The solution is approximately:

`\theta_(2) \approx 22.61^(\textdegree)`