Question

"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are released from rest from the same height H above the horizontal ground. Ball A falls straight down (no air resistance) while B rolls down the inclined plane without slipping. Which ball has the greater TME at the bottom of the incline?

Answer 1

he sphere that uses less time is sphere A

Step-by-step explanation:

Let's start with ball A, for this let's use the kinematics relations

v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

v² = 0 - 2 g (0- y₀)

v =
`√(2g y_o)`

v =
`√(2 \ 9.8\ H)`

v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

Em₀ = U = m g y

final point. At the bottom of the ramp

Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mg H = ½ m v² + ½ I w²

angular and linear velocity are related

v = w r

w = v / r

the momentorot of inertia indicates that it is worth

I =
`(2)/(5)` m r²

we substitute

m g H = ½ m v² + ½ (
`(2)/(5)` m r²) (
`(v)/(r)` )²

gH =
`(1)/(2)` v² +
`(1)/(5)` v² =
`(7)/(10)` v²

v =
`\sqrt{(10)/(7) \ g H}`

v =
`\sqrt{ (10)/(7) \ 9.8 \ H}`

v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H. The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

sin θ = H / L

L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A