Question

A 750 g disk and a 760 g ring, both 15 cm in diameter, are rolling along a horizontal surface at 1.7 m/s when they encounter a 15 ? slope.

How far up the slope does the disk travel before rolling back down?

How far up the slope does the ring travel before rolling back down?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the disk = M₁ = 750 g = 0.75 kg

Mass of the ring = M₂ = 760 g = 0.76 kg

Diameter of the disk = Diameter of the ring = D = 15 cm = 0.15 m

Radius of the disk = Radius of the ring = R = D/2 = 0.075 m

Speed of the disk at the bottom of the slope = Speed of the ring at the bottom of the slope = V = 1.7 m/s

Angle of slope = = 15^o

Moment of inertia of the disk = I₁

Moment of inertia of the ring = I₂

I₁ = M₁R²/2

I₁ = (0.75)(0.075)²/2

I₁ = 2.109 x 10^-3 kg.m²

I₂ = M₂R²

I₂ = (0.76)(0.075)²

I₂ = 4.275 x 10^-3 kg.m²

Angular speed of the disk at the bottom of the slope = ₁

₁ = V/R = 1.7/0.075 = 22.67 rad/s

Angular speed of the ring at the bottom of the slope = ₂

₂ = V/R = 1.7/0.075 = 22.67 rad/s

Height gained by the disk before rolling back down = h₁

Length traveled by the disk up the slope before rolling back down = L₁

h₁ = L₁Sin

Height gained by the ring before rolling back down = h₂

Length traveled by the ring up the slope before rolling back down = L₂

h₂ = L₂Sin

The total kinetic energy of the disk at the bottom of the slope is converted into potential energy of the disk before it starts rolling back down.

M₁V²/2 + I₁₁²/2 = M₁gh₁

(0.75)(1.7)²/2 + (2.109x10^-3)(22.67)²/2 = (0.75)(9.81)L₁Sin(15)

L₁ = 0.854 m

The total kinetic energy of the ring at the bottom of the slope is converted into potential energy of the ring before it starts rolling back down.

M₂V²/2 + I₂₂²/2 = M₂gh₂

(0.76)(1.7)²/2 + (4.275x10^-3)(22.67)²/2 = (0.76)(9.81)L₂Sin(15)

L₂ = 1.138 m

The disk travels 0.854 m up the slope before rolling back down.

The ring travels 1.138 m up the slope before rolling back down.