Question 311: Work, Energy and Power & Circular Motion

Question

A student skates up a ramp that makes an angle 30$$^\circ$$ with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v
0
and wants to turn around over a semicircular path xyz of radius R during which he/she reaches a maximum height h (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)

Accepted Answer

Apply energy conservation,
$${1 \over 2}mv_0^2 = mgh + {1 \over 2}m{v^2}$$
$${1 \over 2}m{v^2} = {1 \over 2}mv_0^2 - mgh$$ ..... (i)
At top point y, the centripetal force provided by the component of mg,
mg sin$$	heta$$ = centripetal force = $${{m{v^2}} \over R}$$
$${{m{v^2}} \over R} = mg\sin 	heta $$
Putting this value in Eq. (i), we get
$${{mgR\sin 	heta } \over 2} = {1 \over 2}mv_0^2 - mgh$$
$${{mgR} \over 2}{1 \over 2} = {1 \over 2}mv_0^2 - mgh$$
$${{gR} \over 2} = v_0^2 - 2gh$$
At point x and a of circular path, the points are at same height but less than h. So the velocity is more than point y.
So required centripetal force = $$\frac{mv^{2}}{r}$$ is more.

Answer

$$v_0^2 - 2gh = {1 \over 2}gR$$

Answer

$$v_0^2 - 2gh = {{\sqrt 3 } \over 2}gR$$

Answer

the centripetal force required at points x and z is zero

Answer

the centripetal force required is maximum at points x and z

Work, Energy and Power & Circular Motion

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