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 

2 Answers

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  • NCS
    Lv 7
    2 months ago
    Favourite 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

    If you find this helpful, please select Favorite Answer. You get points too!

    Attachment image
    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

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