Doubt..Help

Let $M_B$ be the mass of the box, and let $M_W$ be the combined mass of the wheels. Let $x$ be the distance the cart has rolled down the ramp, and let $R$ be the radius of the wheels. We are not told the exact shape of the wheels, so we introduce a parameter $\alpha$ such that the combined moment of inertia of the wheels is $\alpha M_w R^2$ . As the cart rolls down the ramp, both the box and the wheels translate, and the wheels rotate as well. Conversion of gravitational potential energy to kinetic energy results in the following equation:

$\frac{1}{2} (M_B + M_W) \dot{x}^2 + \frac{1}{2} (\alpha M_W R^2) \Big( \frac{\dot{x}}{R} \Big)^2 = (M_B + M_W) g \, x \sin \theta$

Simplifying gives:

$\dot{x}^2 = \frac{(M_B + M_W) g \, x \sin \theta}{\frac{1}{2} (M_B + M_W) + \frac{1}{2} (\alpha M_W) }$

It is likely that the wheels are either disks $(\alpha = \frac{1}{2})$ or circular rims $(\alpha = 1)$ . Using the above expression for the speed of the cart as a function of the cart position, numerically integrate the position of the cart using both $\alpha$ values and choose the value of $\alpha$ which corresponds to $x = 15$ at $t = 3$ . For this simulation, set $M_B = M_W = 1$ . As it turns out, the wheels are circular rims $(\alpha = 1)$ .

Now that we know $\alpha$ , we can double the value of $M_B$ and run the simulation again and see at what time $x = 15$ . The final $t$ value turns out to be $2.828$ . Simulation code is attached. Note that this problem could be done fully analytically (by hand), but it was more convenient for me to use numerical integration.

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import math

theta = math.pi/6.0

m = 1.0
g = 10.0

MB = m
MW = m

alpha = 1.0

dt = 10.0**(-6.0)

#######################################

x = 0.0
xd = 10.0**(-6.0)
t = 0.0

while t <= 3.0:

    x = x + xd * dt

    right = (MB + MW)*g*x*math.sin(theta)
    left = 0.5*(MB+MW) + 0.5*alpha*MW

    xd = math.sqrt(right / left)

    t = t + dt

#######################################

print t
print x
print ""

#3.00000000007
#14.9999220657


#######################################

MB = 2.0*m

x = 0.0
xd = 10.0**(-6.0)
t = 0.0

while x <= 15.0:

    x = x + xd * dt

    right = (MB + MW)*g*x*math.sin(theta)
    left = 0.5*(MB+MW) + 0.5*alpha*MW

    xd = math.sqrt(right / left)

    t = t + dt

#######################################

print t

#2.82843500004

In the first case, the force balance equation gives

$2mg\sin α-F_f=2mw$ ,

where $m, g, w, F_f$ are mass of the box, acceleration due to gravity, acceleration of the box and force of friction respectively. The moment balance equation gives

$F_f\times r=\eta mr^2β$ ,

where $r,β,\eta$ are the radius of each wheel angular acceleration of each wheel and the ratio of total moment of inertia of the wheels and $mr^2$ respectively. Since the rolling is pure, therefore

$w=rβ$

Solving we get $w=\dfrac{2g\sin α}{2+\eta}$ .

In the second case, the corresponding equations are

$3mg\sin α-F'_f=3mw'$

and

$F'_f\times r=\eta mr^2β'$ ,

with

$w'=rβ'$ .

Solving we get $w'=\dfrac{3g\sin α}{3+\eta}$ .

If $t$ and $t'$ be the times of travel in the two cases, then, since the initial velocities in both the cases are zero, we have

$\dfrac{t'}{t}=\sqrt {\dfrac{w}{w'}}$ ,

or

$t'=3\times \sqrt {\dfrac{2(3+\eta) }{3(2+\eta) }}$

For the wheels to be in the shape of circular rims, $\eta=1$ , and the time is given by

$t'=3\times \sqrt {\dfrac{8}{9}}\approx \boxed {1.8856}$ sec.

For the wheels to be in the shape of circular disks, $\eta =\dfrac{1}{2}$ , and the time is given by

$t'=3\times \sqrt {\dfrac{14}{15}}\approx \boxed {2.898}$ sec.

For impure rolling, the force balance equations in the two cases are

$2mg\sin α-2\mu_kmg\cos α=2mw\Rightarrow w=g(\sin α-\mu_k\cos α)$

$3mg\sin α-3\mu_kmg\cos α=3mw'\Rightarrow w'=g(\sin α-\mu_k\cos α)=w$ .

where $\mu_k$ is the coefficient of sliding friction.

Hence both the times will be equal in the case of impure rolling. That is the time will be $3$ sec. in the second case also.