# Particle sliding down a hemispherical bowl

• Tanya Sharma
In summary, a particle P and a bead Q are released from point A at the same time, with P having a horizontal velocity of v and Q having a horizontal velocity of v. P slides down a frictionless hemispherical bowl while Q travels along a horizontal string. The horizontal distance traveled by both particles is the same, but due to the presence of a normal force in the horizontal direction, P's horizontal velocity is greater than v. Therefore, it can be inferred that P will reach point B earlier than Q.
Tanya Sharma

## Homework Statement

A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t=0 along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected.Which bead reaches point B earlier?

## The Attempt at a Solution

Let θ be the angle which the particle P makes with the vertical .

N-Mgcosθ = Mv2/R

Mgsinθ = -Mvdv/dx

or,Mgsinθ = -Mvdv/Rdθ

RMgsinθdθ = -Mvdv

β0RMgsinθdθ = -∫uwMvdv , where β is the initial angle ; u is the initial velocity and w is the velocity at the bottommost point .

But how to relate initial velocity 'u' with the initial horizontal component of velocity 'v' .

I am not sure if this is the correct approach.

I would be grateful if someone could help me with the problem.

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Hi Tanya,

N is not constant. You need to do the force balance in the tangential direction, not the radial
direction.

Chet

Use energy conservation

Does either particle have a horizontal component of acceleration?

1 person
Chestermiller said:
Hi Tanya,

N is not constant. You need to do the force balance in the tangential direction, not the radial
direction.

Chet

Hi Chet...

I might have made some mistake , but I think I have written force equation in tangential direction in post#1.Are you trying to convey something else ?

TSny said:
Does either particle have a horizontal component of acceleration?

Hello TSny

here is my reasoning...

There is a component of normal Nsinθ in horizontal direction .Even though the magnitude of this force decreases as the particle slides down ,still it accelerates the particle in horizontal direction till it reaches the bottom .The horizontal component of velocity at any instant during the downward journey ,say vP,x will be greater than 'v' .

During the upward journey,it is the same component of normal force Nsinθ which decelerates the particle , till it reaches the opposite point to the initial point.Here either by conservation of energy or by symmetry we can say that the velocity of the particle will be same as that of the initial point .

So we can infer that the horizontal component of velocity will always be greater than 'v' during the entire journey. vP,x >v . The average horizontal speed of the particle is greater than 'v' .

OTOH in case of Q there is no horizontal acceleration,hence velocity of vQ = v .

The horizontal distances are same in both the cases ,hence tP < tQ

Does this make sense ?

Tanya Sharma said:
here is my reasoning...

There is a component of normal Nsinθ in horizontal direction .Even though the magnitude of this force decreases as the particle slides down ,still it accelerates the particle in horizontal direction till it reaches the bottom .The horizontal component of velocity at any instant during the downward journey ,say vP,x will be greater than 'v' .

During the upward journey,it is the same component of normal force Nsinθ which decelerates the particle , till it reaches the opposite point to the initial point.Here either by conservation of energy or by symmetry we can say that the velocity of the particle will be same as that of the initial point .

So we can infer that the horizontal component of velocity will always be greater than 'v' during the entire journey. vP,x >v . The average horizontal speed of the particle is greater than 'v' .

OTOH in case of Q there is no horizontal acceleration,hence velocity of vQ = v .

The horizontal distances are same in both the cases ,hence tP < tQ

Does this make sense ?

Excellent!

1 person

## 1. What is a particle sliding down a hemispherical bowl?

A particle sliding down a hemispherical bowl refers to the motion of a small object or particle as it moves along the curved surface of a bowl that is shaped like half of a sphere.

## 2. What is the force acting on the particle as it slides down the bowl?

The force acting on the particle is primarily due to gravity. As the particle moves down the curved surface of the bowl, it experiences a downward force due to the pull of gravity towards the center of the Earth.

## 3. How does the speed of the particle change as it slides down the bowl?

The speed of the particle increases as it slides down the bowl due to the acceleration of gravity. The particle will continue to gain speed until it reaches the bottom of the bowl or until it encounters friction or other external forces that may slow it down.

## 4. What factors can affect the motion of the particle as it slides down the bowl?

The motion of the particle can be affected by several factors, including the size and mass of the particle, the shape and angle of the bowl, and the presence of any external forces such as friction or air resistance. The location of the particle on the bowl's surface can also impact its motion.

## 5. How is the motion of the particle related to the curvature of the bowl?

The motion of the particle is directly related to the curvature of the bowl. The steeper the curve of the bowl, the faster the particle will accelerate as it slides down. A flatter bowl will result in a slower acceleration and speed for the particle.

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