A student skates up a ramp that makes an angle 30 ° 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 x y z of radius R during which he/she reache

Question

A student skates up a ramp that makes an angle $30 °$ 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 $x y z$ 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) [2020]

Smart Achievers · Accepted Answer

(1, 4)

At point Y the centripetal force provided by the component of weight mg

$v_{0}$ $m g \sin 30 ° = \frac{m v^{2}}{R}$

$∴$ $v^{2} = \frac{g R}{2}$

Now by the energy conservation between bottom point and point Y

$\frac{1}{2} m v_{0}^{2} = m g h + \frac{1}{2} m v^{2}$

$∴$ $v^{2} = v_{0}^{2} - 2 g h ...(i)$

$∴$ $From eq. (i)$

$\frac{g R}{2} = v_{0}^{2} - 2 g h$

Hence option (1) is correct.

At point x and z of circular path, the points are at same height but less than h. So the velocity is more than at point y.

So required centripetal force $= \frac{m v^{2}}{r}$ is maximum at points x and y.

Solution

1 $v_{0}^{2} - 2 g h = \frac{1}{2} g R$

2 $v_{0}^{2} - 2 g h = \frac{\sqrt{3}}{2} g R$

3 the centripetal force required at points $x$ and $z$ is zero

4 the centripetal force required is maximum at points $x$ and $z$

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Solution

1 v02-2gh=12gR

2 v02-2gh=32gR

3 the centripetal force required at points x and z is zero

4 the centripetal force required is maximum at points x and z

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1 $v_{0}^{2} - 2 g h = \frac{1}{2} g R$

2 $v_{0}^{2} - 2 g h = \frac{\sqrt{3}}{2} g R$

3 the centripetal force required at points $x$ and $z$ is zero

4 the centripetal force required is maximum at points $x$ and $z$