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 v0 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)

the centripetal force required is maximum at points x and z

v_0^2 - 2gh = \sqrt 3 \over 2gR

the centripetal force required at points x and z is zero

Work Power and Energy PYQ — JEE Advanced 2020 · Shift 2 · Q44 with Solution

View written solutionFree

Correct answer: D

Speed at the highest point $y$ using centripetal condition

At the top point $y$ of the semicircular turn, the problem says that the force required for the turn is provided by the student's weight only.

So at $y$ , the only force toward the center is $mg$ .

Hence,

\frac{mv_y^2}{R}=mg

which gives

v_y^2=gR

Apply conservation of mechanical energy from start to point $y$

The student starts at ground level with speed $v_0$ . At point $y$ , the height above ground is $h$ and speed is $v_y$ .

Since energy loss is negligible,

\frac12 mv_0^2 = \frac12 mv_y^2 + mgh

Substitute $v_y^2=gR$ :

\frac12 mv_0^2 = \frac12 m(gR) + mgh

Multiply by $2/m$ :

v_0^2 = gR + 2gh

So,

v_0^2 - 2gh = gR

This does not match option A or B. So A is false and B is false.

Centripetal force required at points $x$ and $z$

The centripetal force required is

F_c = \frac{mv^2}{R}

At points $x$ and $z$ , the height is lower than at $y$ , so by energy conservation the speed is greater there than at $y$ . Thus,

v_x = v_z > v_y

Therefore,

\frac{mv_x^2}{R} = \frac{mv_z^2}{R} > \frac{mv_y^2}{R}

So the centripetal force required is not zero at $x$ and $z$ . Hence C is false.

Also, since speed is maximum at the lowest points of the circular path, the required centripetal force is maximum at $x$ and $z$ . Hence D is true.

Conclusion

The correct option is:

D only

Comparison with stored correct answer

Stored correct answer: A, D

My derived answer: D only

There seems to be a discrepancy. Using the standard condition at the top point,

\frac{mv_y^2}{R}=mg \Rightarrow v_y^2=gR,

which leads to

v_0^2 - 2gh = gR,

not $\frac12 gR$ .

So option A appears inconsistent unless the radius effectively used for curvature at the top is $2R$ instead of $R$ , which is not what the question states.

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View written solutionFree

Correct answer: D

Speed at the highest point $y$ using centripetal condition

At the top point $y$ of the semicircular turn, the problem says that the force required for the turn is provided by the student's weight only.

So at $y$ , the only force toward the center is $mg$ .

Hence,

\frac{mv_y^2}{R}=mg

which gives

v_y^2=gR

Apply conservation of mechanical energy from start to point $y$

The student starts at ground level with speed $v_0$ . At point $y$ , the height above ground is $h$ and speed is $v_y$ .

Since energy loss is negligible,

\frac12 mv_0^2 = \frac12 mv_y^2 + mgh

Substitute $v_y^2=gR$ :

\frac12 mv_0^2 = \frac12 m(gR) + mgh

Multiply by $2/m$ :

v_0^2 = gR + 2gh

So,

v_0^2 - 2gh = gR

This does not match option A or B. So A is false and B is false.

Centripetal force required at points $x$ and $z$

The centripetal force required is

F_c = \frac{mv^2}{R}

At points $x$ and $z$ , the height is lower than at $y$ , so by energy conservation the speed is greater there than at $y$ . Thus,

v_x = v_z > v_y

Therefore,

\frac{mv_x^2}{R} = \frac{mv_z^2}{R} > \frac{mv_y^2}{R}

So the centripetal force required is not zero at $x$ and $z$ . Hence C is false.

Also, since speed is maximum at the lowest points of the circular path, the required centripetal force is maximum at $x$ and $z$ . Hence D is true.

Conclusion

The correct option is:

D only

Comparison with stored correct answer

Stored correct answer: A, D

My derived answer: D only

There seems to be a discrepancy. Using the standard condition at the top point,

\frac{mv_y^2}{R}=mg \Rightarrow v_y^2=gR,

which leads to

v_0^2 - 2gh = gR,

not $\frac12 gR$ .

So option A appears inconsistent unless the radius effectively used for curvature at the top is $2R$ instead of $R$ , which is not what the question states.

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Work Power and Energy question

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