# Can anyone figure out this physics question?

A billiard ball of mass 240 g and radius 3.10 cm starts with a translational speed of 1.00 m/s at point A on the track as shown in the figure below. If point B is at the top of a hill that has a radius of curvature of 75 cm, what is the normal force acting on the ball at point B? Assume the billiard ball rolls without slipping on the track.

the figure shows that the ball has a height difference of 10 cm between the two points.

Thank you!

I thought to use conservation of energy to find velocity and then use the equation N= (mv^2)/r + mg to solve for N. But I can't tell what I am doing wrong

### 2 Answers

- NCSLv 72 months agoFavourite answer
I'll assume your diagram looks like the one below.

I'll further wager a guess that you're not considering the energy of rotation when you find the velocity at B.

total E = KEtrans + KErot = ½mv² + ½Iω²

For a solid sphere, I = (2/5)mr²

and "without slipping" means ω = v/r

so

E = ½mv² + ½(2/5)mr²(v/r)² = ½mv² + (1/5)mv² = 0.7mv²

so at B,

0.7mv² = m*9.8m/s²*0.10m + 0.7m*(1m/s)²

mass m cancels

v² = 1.4m²/s² + 0.7m²/s² = 2.1 m²/s²

NOW try

N = mg - mv²/r

using g = +9.8 m/s²

I get around 1.5 N

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Source(s): image from study.com - FiremanLv 72 months ago
The initial KE at point A will be = KE(t) + KE(r)

=>KE(A) = 1/2mv^2 + 1/2Iω^2

=>KE(A) = 1/2mv^2 + 1/2 x 2/5mr^2 x (v/r)^2 {as I = 2/5mr^2 & v = rω}

=>KE(A) = 1/2mv^2 + 1/5mv^2

=>KE(A) = 7mv^2/10 -----------(i)

& PE(A) = mgh

Let the linear velocity at B is u m/s^2

=>KE(B) = 7mu^2/10 by (i)

& PE(B) = mg(h -0.1)

Thus by the law of energy conservation:

=>KE(A) +PE(A) = PE(B) + KE(B)

=>7mv^2/10 + mgh = 7mu^2/10 + mg(h - 0.1)

=>7v^2/10 + gh = u^2/10 + gh - 0.1g

=>u^2 = 7v^2 + g

=>u^2 = 7 x (1)^2 + 9.8

=>u = √16.8

=>u = 3.07 m/s

In the absence of figure it is assumed that A is 10 cm above B. However N may be calculated by N = mu^2/r +/- mg, the +/- sign can not be decided in the absence of figure.

But it may have two solutions i.e.

By N = mu^2/r +/- mg

=>N = m[(3.07)^2/0.75 +/- 9.8]

=>N = 0.24 x [12.57 +/- 9.8]

=>N for +ve = 5.37 N

& N for -ve = 0.66 N

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