Question

The radius of a wheel is 0.680 m. A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude 5.00 N, unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope.What is the angular displacement Δθ, in radians, of the wheel during 1.20 revolution?

Answer 1

Givens.

• The radius of the wheel is 0.680 m.

,

• The applied force is 5.00 N, and it's a tangential force.

,

• The number of revolutions is 1.20.

We need to find the circumference of the wheel, use the circumference formula.

$C=2\pi r$

Where r = 0.680 m and pi = 3.14.

$C=2\cdot3.14\cdot0.680=4.27m$

The circumference of the wheel is 4.27 meters.

It's important to notice that the circumference represents 1 revolution, so for 1.20 revolutions we have

$1.20\text{rev}\cdot\frac{4.27m}{1\text{rev}}=5.12m$

This means the total distance covered is 5.12 m. However, the displacement is different because it takes into account only the change of position, so basically, the displacement would be 0.20 revolutions because at 1 revolution we are at the same starting point.

$0.20\text{rev}\cdot\frac{4.27m}{1\text{rev}}=0.854m$

Therefore, the total displacement is 0.854 meters.

However, we need to find the angular displacement, so

$\theta=\frac{d}{r}=\frac{0.854m}{0.680m}=1.26\text{rad}$

Hence, the angular displacement is 1.26 radians.