Gerard asked in Science & MathematicsPhysics · 2 months ago

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.

Update:

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

Thank you!

Update 2:

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

Relevance
• NCS
Lv 7
2 months ago

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

Source(s): image from study.com
• 2 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