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.A. How much rope unwinds while the wheel makes 1.00 revolution? (m)B. How much work is done by the rope on the wheel during 1.00 revolution? (J)C. What is the torque on the wheel about its axis due to the rope? (N*m)D. What is the angular displacement Δθ, in radians, of the wheel during 1.20 revolution? (rad)

Answer 1

Given data

*The given radius of the wheel is r = 0.680 m

*The given force of magnitude is F = 5.00 N

*The given angular displacement is

`\Delta\theta=1\text{ rev= 2}\pi\text{ radian}`

(A)

In one revolution, a length equals the circumference of rope winds.

`\begin{gathered} C=2\pi r \\ =2*3.14*0.680 \\ =4.27\text{ m} \end{gathered}`

(B)

The formula for the work done by the rope on the wheel during the 1.00 revolution is given as

`\begin{gathered} W=\tau\Delta\theta \\ =(F\mathrm{}r)\Delta\theta \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} W=(5.00*0.680)(2\pi) \\ =21.35\text{ J} \end{gathered}`

Hence, the work is done by the rope on the wheel during the 1.00 revolution is W = 21.35 J

(C)

The formula for the torque on the wheel about its axis due to the rope is given as

`\tau=r* F`

Substitute the known values in the above expression as

`\begin{gathered} \tau=(5.00)(0.680) \\ =3.4\text{ N.m} \end{gathered}`

Hence, the torque on the wheel about its axis due to the rope is 3.4 N.m

(D)

The angular displacement in radians, of the wheel during the 1.20 revolution is calculated as