Question

A solid 0.6350 kg ball rolls without slipping down a track toward a vertical loop of radius R=0.8950 m.

What minimum translational speed vmin must the ball have when it is a height H=1.329 m above the bottom of the loop in order to complete the loop without falling off the track?

Assume that the radius of the ball itself is much smaller than the loop radius R. Use g=9.810 m/s2 for the acceleration due to gravity.

Answer 1

The minimum translational speed the ball must have at a height of 1.329 m to complete the loop without falling off the track is approximately 6.61 m/s.

How can you solve the minimum translational speed the ball must have?

E p(top) = K e(bottom)

E p(top) = m * g * (R + H)

where:

m is the mass of the ball (0.6350 kg)

g is the acceleration due to gravity (9.810 m/s²)

R is the radius of the loop (0.8950 m)

H is the height above the bottom of the loop (1.329 m)

Calculate the minimum kinetic energy at the bottom:

Since the ball needs enough speed at the bottom to reach the top again, the minimum kinetic energy at the bottom is equal to the potential energy at the top:

K e(bottom) = E p(top) = m * g * (R + H)

Find the minimum translational speed:

K e = 1/2 * m * vmin²

where vmin is the minimum translational speed we're looking for. Solving for vmin:

v min = √(2 * K e / m) = √(2 * m * g * (R + H) / m)

v min = √(2 * g * (R + H))

Plug in the values and calculate:

v min = √(2 * 9.810 * (0.8950 + 1.329))

v min ≈ 6.61 m/s

Therefore, the minimum translational speed the ball must have at a height of 1.329 m to complete the loop without falling off the track is approximately 6.61 m/s.

Answer 2

The minimum translation speed is 4.21 m/s.

Step-by-step explanation:

Given that,

Mass of solid = 0.6350 kg

Radius = 0.8950 m

Height = 1.329 m

We need to calculate the speed of the ball

Using formula of centripetal force

`F=(mv^2)/(r)`

`mg=(mv^2)/(r)`

`v_(b)=√(rg)`

Put the value into the formula

`v_(b)=√(0.8950*9.8)`

`v_(b)=2.961\ m/s^2`

We need to calculate the minimum translation speed

Using conservation of energy

`K.E_(i)+P.E_(i)=K.E_(f)+P.E_(f)`

`(1)/(2)mv_(i)^2+mg(H-2r)=(1)/(2)mv_(f)^2+0`

`v_(i)^2=\sqrt{v_(f)^2-(2g(H-2r))}`

Put the value into the formula

`v_(i)^2=(2.961)^2-(2*9.8(1.329-2*0.8950))`

`v_(i)^2=√((2.961)^2-(2*9.8(1.329-2*0.8950)))`

`v_(i)=4.21\ m/s`

Hence, The minimum translation speed is 4.21 m/s.