Question

1. A 4.8 kg weight, mw, slides down a frictionless ramp that is at an angle of  = 22 with respect
to the horizontal. The weight is attached to a massless string, which in turn is threaded over a
pulley (solid disk) of radius rp = 4.5 cm and mass mp = 0.55 kg. The string is then wound around
a thin hollow metal cylinder (hoop) of radius rc = 9.4 cm and mass mc = 8.6 kg. Assuming the
wheel starts from rest, find its velocity after it has traveled a distance s = 15 cm down the ramp.

Answer 1

Given that,

Mass of weight = 4.8 kg

Angle = 22°

Radius of pulley= 4.5 cm

Mass of pulley = 0.55 kg

Radius of cylinder = 9.4 cm

Mass of cylinder = 8.6 kg

Distance = 15 cm

We need to calculate the velocity after it has traveled a distance

Using conservation of energy

$m_{w}g\sin\theta\times d=\dfrac{1}{2}I_{p}\omega_{p}^2+\dfrac{1}{2}I_{c}\omega_{c}^2+\dfrac{1}{2}m_{w}v^2$

$m_{w}g\sin\theta\times d=\dfrac{1}{2}\times\dfrac{1}{2}m_{p}r_{p}^2\omega_{p}^2+\dfrac{1}{2}m_{c}r_{c}^2\omega_{c}^2+\dfrac{1}{2}m_{w}v^2$

$m_{w}g\sin\theta\times d=\dfrac{1}{2}(\dfrac{1}{2}m_{p}v^2+m_{c}v^2+m_{w}v^2)$

Put the value into the formula

$4.8\times9.8\times\sin22\times0.15=\dfrac{1}{2}(\dfrac{1}{2}\times0.55+8.6+4.8)v^2$

$2.707=6.8375v^2$

$v^2=\dfrac{2.7078}{6.8375}$

$v=\sqrt{\dfrac{2.7078}{6.8375}}$

$v=0.629\ m/s$

Hence, The velocity after it has traveled a distance is 0.629 m/s.