A roller coaster car moves in a vertical circle of radius R. At the top of the circle the car has speed v1, and at the bottom of the circle it has speed v2, where v2>v1. Suppose that v2 is in the x-direction, and y-axis is directed upward.

1) When the car is at the top of its circular path, what is the direction of its radial acceleration, arad,top? The direction angle β-top of the radial acceleration arad,top is measured from the +x-direction toward the +y-direction.

2) When the car is at the bottom of its circular path, what is the direction of its radial acceleration, arad,bottom? The direction angle β-bottom of the radial acceleration arad,bottom is measured from the +x-direction toward the +y-direction.

3) In terms of v1 and v2, what is the ratio arad,bottom/arad,top?

1 ) Radial acceleration is nothing but centripetal acceleration . When the car is at the top , centripetal acceleration is downwards towards the center.

In other words , its direction is at 270 degree from + x axis .

angle β-top = 270 degree

2 ) When the car is at the lowest position , centripetal acceleration is again

towards the center or in upward direction .

So angle β-bottom = 90 degree.

3 ) Velocity at the bottom v₂ = √5gr

velocity at the top v₁ = √gr

v₂ / v₁ = √5 .

OR

arad(bottom ) = v₂² / R

arad( top) = v₁² / R

arad(bottom ) /arad( top) = v₂² / v₁²

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-01-25T13:02:20+01:00

The value of β-top = 3π/2 and β-bottom = π/2 and the ratio of radial acceleration at the bottom over the radial acceleration at the top is 5:1.

Given that the speed of the roller coaster car at the top of the vertical circle is v₁ and at the bottom of the vertical circle is v₂.

When the car is at the bottom its velocity is in x-direction, v₂ is in +ve x-direction.
Which implies that v₁ will be in -ve x-direction, since at the top the tangential velocity will be opposite to the tangential velocity at the bottom.
Also the +ve y -direction is diected upwards.

(1) when the car is at the top:

The radial acceleration is always directed towards the centre.

So the radial acceleration is directed downwards, in -ve y-direction.

If β-top be the angle of radial acceleration with respect to +ve x-direcion, then, since the radial acceleratiion points in -ve y-direction.

β-top = 3π/2

(2) When the car is at the bottom

The radial acceleration is always directed towards the centre.

So in this case the radial acceleration is directed upwards, in +ve y-direction.

If β-bottom be the angle of radial acceleration with respect to +ve x-direcion, then, since the radial acceleratiion points in +ve y-direction.

β-bottom = π/2

(3) To execute vertical circular motion, the velocities at the top and at the bottom are:

[tex]v_{top}=\sqrt{gr} = v_1\\\\v_{bottom}=\sqrt{5gr}=v_2[/tex]

now the radial acceleration is given by:

[tex]a_r=v^2/r\\ \\ a_{bottom}= {v_2}^2/r\\\\ a_{top}={v_1}^2/r\\\\ \frac{a_{bottom}}{a_{top}}= \frac{(v_2)^2}{(v_1)^2}\\ \\ \frac{a_{bottom}}{a_{top}}= 5[/tex]

Hence, the ratio [tex]a_{bottom}:a_{top}=5:1[/tex]

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