Question

A large 3kg object hangs from a rope wound on a 40 kg wheel. The wheel has an actual radius of 0.75m and a radius of gyration of 0.60m. Find the angular acceleration and the distance through which the weight will fall in the first 10s.

Answer 1

Answer on Question55195 - Physics / Mechanics — Kinematics — Dynamics - for completion

October 1, 2015

A large $m=3kg$ object hangs from a rope wound on a $M=40kg$ wheel. The wheel has an actual radius of $R=0.75m$ and a radius of gyration of $\rho=0.60m$. Find the angular acceleration and the distance through which the weight will fall in the first $t_{1}=10s$.

Solution

If $a$ is the acceleration if an object, and $T$ is the tension of the rope, then according to the Newton’s second law:

$ma=mg-T$

The moment of the force acting on the wheel:

$M=TR$

The equation of motion of the wheel:

$I\varepsilon=M$

where $I$ is moment of inertia:

$I=M\rho^{2}$

$a=\varepsilon R$

From the equations above:

$M\rho^{2}\varepsilon=m(g-\varepsilon R)R$

$\varepsilon=\frac{mg}{M\rho^{2}+mR^{2}}\approx 1.83rad/s^{2}$

The distance through which the weight will fall in the first 10s:

$d=\frac{at_{1}^{2}}{2}=\frac{\varepsilon Rt_{1}^{2}}{2}\approx 86.6m$

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