Question

As part of an experiment in physics lab, small metal ball of radius r = 2.3 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 3.4 m. The ball is solid with a uniform density and a mass M = 373 g.

How high above the top of the loop must it be released in order that the ball just makes it around the loop?

Now instead of a sphere, what if we rolled a solid disk with the same mass and radius. How high above the top of the loop must it be released in order that the disk just makes it around the loop?

Finally, what if we had allowed a block with the same mass as the sphere to slide on the ramp. The block slips without friction but does not rotate. How high above the top of the loop must it be released in order that the block just makes it around the loop?

Answer 1

Given,

masss of the metal ball, m = 373 g = 0.373 kg

Radius of the metal ball, r = 2.3 cm = 0.023 m

Radius of the loop-the-loop, R = 3.4 m

moment of inertia of sphere, I = 2/5*m*r²

Now,

At the top,

N + mg = mv²/R

For ball to just make it around the loop, N =0,

=> mg = mv²/R

=> v =

Now,

Let angular velocity be ,

As we know,

v = r

=> = v/r

Now,

Let the height at which ball is released be h and initial velocity is zero

Applying conservation of energy, at the starting point and top of the loop

=> mgh = mg*2R + 1/2*m*v² + 1/2*I*²

put value of I and ,

=> mgh = mg*2R + 1/2*m*v² + 1/2*(2/5*m*r²)*(v/r)²

=> mgh = mg*2R + 1/2*m*v² + 1/5*m*v²

=> gh = g*2R + 7/10*v²

Now,

v² = Rg

=> gh = 2gR +7/10*gR

=> h = 2R +7/10*R

Now,

height of the loop is 2R ,

Height above the loop the ball must be released, y = 7/10*R = 7/10*3.4

                                                                                    = 2.38 m

Now,

moment of inertia of solid disk, I₁ = 1/2*m*r²

Now,

Let the height be h₁, and applying conservation of energy

mgh₁ = mg*2R + 1/2*m*v² + 1/2*I*²

put value of I and ,

=> mgh₁ = mg*2R + 1/2*m*v² + 1/2*(1/2*m*r²)*(v/r)²

=> gh₁ = 2Rg + 1/2*m*v² + 1/4*m*v²

Now, v² = Rg

=> gh₁ = 2Rg + 1/2*gR + 1/4*gR

=> h₁ = 2R + 3/4*R,

Since, height of the loop is 2R,

So height above the loop, y₁ = 3/4*R = 0.75*3.4

                                               = 2.55 m

Now,

if disk is released, so there will be no rotational energy,

Let height be h₂

=> mgh₂ = mg*2R + 1/2*m*v²

since, v² = Rg

=> mgh₂ = mg*2R + 1/2*m*Rg

=> h₂ = 2R + 1/2*R

Height above the loop, y₃ = 1/2*R = 0.5 *3.4

                                           = 1.7 m