Answer to Question #200974 in Mechanics | Relativity for Michelle Tammi

Question

A wheel with a radius of 1.961m and a mass of 0.296kg rolls without sliding down a plane that tilts at an angle of 21.940 degrees. If the wheel is accelerated by a torque of 1.595 N x m, calculate the power if the wheel, which is shaped like a homogeneous cylinder, has a kinetic energy of 13.90J

Answer 1

A wheel with a radius of 1.961m and a mass of 0.296kg rolls without sliding down a plane that tilts at an angle of 21.940 degrees. If the wheel is accelerated by a torque of 1.595 N x m, calculate the power if the wheel, which is shaped like a homogeneous cylinder, has a kinetic energy of 13.90J

$I\varepsilon=M=F_{fr}r$

where I is moment of inertia of the wheel,

$\varepsilon$ is angular acceleration,

M is torque.

$ma=mgsin\alpha-F_{fr}$

For homogeneous cylinder:

$I=mr^2/2$

For moving without sliding:

$a=\varepsilon r$

The kinetic energy:

$T=\frac{mv^2}{2}+\frac{I\omega^2}{2}$

$\omega=v/r$

Then:

$F_{fr}=M/r$

$F=ma=mgsin\alpha-M/r$

$v=r\sqrt{\frac{2T}{mr^2+I}}=2\sqrt{\frac{T}{3m}}$

Power:

$P=Fv=2(mgsin\alpha-M/r)\sqrt{\frac{T}{3m}}$

$P=2(0.296\cdot9.8sin21.94\degree-1.595/1.961)\sqrt{\frac{13.9}{3\cdot0.296}}=2.14\ W$