Question

A toy cart consists of a 1.00 kg block and four solid rubber wheels 200 g each. The cart is released from rest at the top of an inclined plane at a height of 46.0 cm above the ground, as shown. Determine the speed near the bottom of the incline when a. the cart is placed on its back (wheels up) and slides down without friction. the cart rolls down without slipping. 46 cm

refer to pic

Answer 1

Let M = mass of the car

m = mass of each tyre

Moment of inertia I of each tyre = (1/2)mR^{​​​​​​2}

a) When the cart is sliding ,

Total energy E_{​​​​1} at the top of the inclined plane = K.E + P.E.

E₁ = 0 + (M+4m)gh

{ As initial velocity is zero so initial kinetic energy of the cart is zero }

E₁ = (M+4m)gh

Total energy E₂ at the bottom of the inclined plane = K.E. + P.E.

E_​​2 = 0 + (1/2)(M+4m)v²

Using conservation of energy

E₁ = E₂

(M+4m)gh = (1/2)(M+4m)v²

v = (2gh)^1/2

v = (2×9.8×0.46)^1/2

v = 3 m/s

b) When the cart is rolling,

Let v' be the speed of the cart at the bottom of the plane and w' be the angular speed of each wheel

Total energy of the cart E₁' at the top of the inclined plane = (M+4m)gh

Total energy at the bottom of the inclined plane

= (1/2)×(M+4m)v'² + 4{ (1/2)Iw'² }

= (1/2)×(M+4m)v'² + 4{(1/4)mR^{​​​​​​2}w'²}

= (1/2)×(M+4m)v'² + mv'²    {As v' = w'R}

= (1/2)×(M+6m)v'²

Using conservation of energy,

(M+4m)gh = (1/2)×(M+6m)v'²

v'² = 2(M+4m)gh / (M+6m)

v' = { 2×(1+0.8)×9.8×0.46 / (1+1.2) }^1/2

v' = ( 16.23 / 2.2 )^1/2

v' = (7.38)^1/2

v' = 2.717 m/s