Question

A 150.0 kg cart rides down a set of tracks on four solid steel wheels, each with radius 20.0 cm and mass 45.0 kg. The tracks slope downward at an angle of 17 ∘ to the horizontal. If the cart is released from rest a distance of 12.5 m from the bottom of the track (measured along the slope), how fast will it be moving when it reaches the bottom? Assume that the wheels roll without slipping, and that there is no energy loss due to friction.

Answer 1

Sol::

Given:
Cart mass (M) = 150.0 kg
Number of wheels= 4
radius(r)= 0.2 m
mass(m)= 45.0 kg.
angle Ѳ = 17°
distance(d)= 12.5 m

As per given data we have

Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom
i .e

(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²)

Where
The height (h) will be
sin 17° = h / 12.5
h = 3.654 m
And
For disks about the z-axis,we have
I = mr² / 2
So

(M+4m)gh = 0.5*(M+4m)v² + 4(0.5(mr²/2) ω²)
(M+4m)gh = 0.5 *(M+4m)v² + m r² ω²
As ω= v/r
(M+4m)gh = 0.5*(M+4m)v² + m v²
(M+4m)gh = v²(0.5M+2m+m)
v² =(M+4m)gh /(0.5M+3m)
v = √[(150+4*45)*9.8*3.65/(0.5*150+3*45)]
v = 7.497 m/s

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