A student skates up a ramp that makes an angle 30 degrees with the horizontal. He/she starts (as sho

Q: Why do we use $mg \sin 30^\circ$ and not just $mg$ for the centripetal force?

The circular path lies *on* the inclined plane. The force required to keep the student in this circle must also lie on the plane. The component of gravity acting along the plane is $mg \sin 30^\circ$. The perpendicular component $mg \cos 30^\circ$ is balanced by the normal force.

Q: Why is the velocity at $x$ and $z$ greater than at $y$?

Points $x$ and $z$ are lower on the ramp than point $y$. According to the conservation of energy, a lower potential energy means a higher kinetic energy, which translates to a higher speed.

Q: Does friction play a role in this problem?

The problem explicitly states that at the highest point, the force required for the turn is provided by weight *only*. This implies we don't need to consider friction for that specific calculation. Energy loss is also stated to be negligible.

Q: How do we know the center of the semicircle is directly down the slope from $y$?

Point $y$ is defined as the maximum height of the semicircular path. On an inclined plane, the highest point of a circle is the point furthest 'up' the slope. Therefore, the center of the circle must lie directly 'down' the slope from this apex.

Q: Why is the centripetal force maximum at $x$ and $z$?

The required centripetal force is given by $F_c = \frac{mv^2}{R}$. Since the radius $R$ and mass $m$ are constant, the force is directly proportional to the square of the speed. Because the speed is maximum at the lowest points ($x$ and $z$), the required force is also maximum there.

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)

Options:

(A) $v_0^2 - 2gh = \frac{1}{2}gR$

(B) $v_0^2 - 2gh = \frac{\sqrt{3}}{2}gR$

(C) the centripetal force required at points x and z is zero

(D) the centripetal force required is maximum at points x and z

Sigma Percentile · Accepted Answer

### The Setup: A 3D Dance of Energy

Imagine you are standing at the bottom of a massive ramp, inclined at exactly $30^\circ$ to the horizontal. You push off with an initial speed $v_0$, feeling the rush of wind as you glide upwards. But you don't just go straight up; you carve a beautiful, sweeping semicircular path across the face of the incline. This path, labeled $xyz$, has a perfect radius $R$.

As you reach the absolute apex of this turn, point $y$, you are at a maximum height $h$ above the solid ground below. This isn't just a geometry problem; it's a dynamic dance between your kinetic energy and the relentless pull of gravity.

### The Master Equation: Conservation of Energy

The problem gives us a beautiful gift: 'energy loss is negligible'. This phrase is our golden ticket to use the Principle of Conservation of Mechanical Energy.

At the very bottom of the ramp, all your energy is kinetic. You are moving fast, and your height is zero.

$$E_{	ext{initial}} = \frac{1}{2}mv_0^2$$

As you climb to point $y$, gravity does negative work on you, converting that precious kinetic energy into gravitational potential energy. At height $h$, your potential energy is $mgh$, and you still have some leftover kinetic energy because you are still moving across the ramp with a velocity $v_y$.

$$E_{	ext{final}} = mgh + \frac{1}{2}mv_y^2$$

Equating the two, we get our master equation:

$$\frac{1}{2}mv_0^2 = mgh + \frac{1}{2}mv_y^2$$

Let's isolate the velocity term, as it will be crucial for our next step:

$$v_y^2 = v_0^2 - 2gh$$

### The Apex: Dynamics at Point y

Now, let's shift our perspective. Imagine looking at the ramp from a top-down view, perpendicular to the inclined surface. From this angle, your path $xyz$ is a perfect semicircle.

At point $y$, you are at the very top of this semicircle. To keep you moving in this circular arc, the universe demands a centripetal force directed towards the center of the circle, $C$.

$$F_c = \frac{mv_y^2}{R}$$

But where does this force come from? The problem states a critical constraint: 'the force required for this turn at the highest point is provided by his/her weight only.'

### The Crucial Constraint: Weight as the Restoring Force

Think about the forces acting on you at point $y$. Gravity pulls straight down towards the center of the Earth. But we are on an inclined plane! We must resolve gravity into two components: one perpendicular to the ramp ($mg \cos 30^\circ$), which is balanced by the normal force, and one parallel to the ramp, pointing straight down the slope ($mg \sin 30^\circ$).

Because point $y$ is the highest point on the semicircular path, the center of the circle $C$ lies exactly straight down the slope from $y$. This means the component of gravity acting down the slope is perfectly aligned to act as our centripetal force!

**This is the crux of the problem.** We can equate the required centripetal force to the available gravitational force component:

$$mg \sin 30^\circ = \frac{mv_y^2}{R}$$

### The Final Calculation

We know that $\sin 30^\circ = \frac{1}{2}$. Substituting this in, we see something beautiful happen—the mass $m$ cancels out entirely!

$$mg \left(\frac{1}{2}ight) = \frac{mv_y^2}{R}$$

$$v_y^2 = \frac{gR}{2}$$

Now, we bring back our master energy equation from earlier:

$$v_y^2 = v_0^2 - 2gh$$

Substitute our new expression for $v_y^2$:

$$\frac{gR}{2} = v_0^2 - 2gh$$

Rearranging this to match the options, we get:

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

This perfectly matches option (A).

### The Extremes: Points x and z

But we aren't done yet. We must evaluate the other options, which ask about the centripetal force at points $x$ and $z$.

Points $x$ and $z$ are the entry and exit points of the semicircular turn. Geometrically, they sit lower on the inclined plane than the apex $y$.

Because they are at a lower height, their gravitational potential energy is less than at $y$. According to our trusty conservation of energy, if potential energy decreases, kinetic energy must increase.

Therefore, the speed at $x$ and $z$ is strictly greater than the speed at $y$:

$$v_x = v_z > v_y$$

The required centripetal force at any point on the circle is $F_c = \frac{mv^2}{R}$. Since the mass $m$ and radius $R$ are constant, the required force scales with the square of the velocity.

Because the velocity is maximum at the lowest points of the turn ($x$ and $z$), **the required centripetal force is also maximum at points $x$ and $z$.**

This confirms that option (D) is absolutely correct.

By systematically breaking down the 3D geometry into energy conservation and 2D circular dynamics, we've conquered a classic JEE Advanced problem. Keep visualizing, keep questioning, and the physics will always reveal its secrets!

Select Answer:

Visualized Solution

Analyzing the Setup

Conservation of Mechanical Energy

Centripetal Force at Apex

Weight Component as $F_{c}$

Evaluating the Expression

Centripetal Force at $x$ and $z$

The Sigma Insight: Conservation of Mechanical Energy

The Setup

The Master Equation

The Apex

The Crucial Constraint

The Final Calculation

The Extremes

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