Atennis ball is a hollow sphere with a thin wall. it is set rolling without slipping at 4.03 m/s on the horizontal section of a track, as shown. it rolls around the inside of a vertical circular loop 90.0 cm in diameter and finally leaves the track at a point 20.0 cm below the horizontal section. (a) find the speed of the ball at the top of the loop. demonstrate that it will not fall from the track. (b) find its speed as it leaves the track. (c) suppose that static friction between the ball and the track was negligible, so that the ball slide instead of rolling. would its speed then be higher, lower, or the same at the top of the loop

2.38 m/s, 4.31 m/s, lower

Explanation:

a)

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

b)

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

c)

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

Refer the below Solution for better understanding.

Given :

Speed = 4.03 m/sec

Vertical circular loop of 90 cm diameter.

Solution :

a)

Initial energy = Final Energy

here,

for a hollow sphere,

by further solving above equation,

 --- (1)

Now put the values of in equation (1),

Now,

F = ma

  --- (2)

Now put the values of R, g, m and in equation (2) we get,

N = 2.77m

ball stays on the track.

b) To find the speed of the ball as it leaves the track,

---- (3)

put h = -0.2m in equation (3)

c) Again, but without rotational energy

Initial energy = Final energy

by further solving the above equation we get,

and this is less than we calculated earlier.

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