91Ó°ÊÓ

We have the given data:

Diameter of the uniform cylinder (D) = 0.300 m

Radius of the uniform cylinder (R) =0.150 m.

Mass of the water bucket (m) =15.0 Kg.

Mass of the uniform cylinder (M) = 12 Kg.

Bucket falling at the height (h) =10 m.

At rest, the velocity $\left(v_1\right)$= 0 , time t = 0

If a particle of mass experiences a nonzero net force, its acceleration is related to the net force by Newton’s Second Law:

$\underset{}{∑}F=ma.......\left(1\right)$

If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object undergoes an angular acceleration , where,

$\underset{}{∑}{\mathrm{Ï„}}_{ext}=l\mathrm{Î\pm }.......\left(2\right)$

When a rigid object rotates about a fixed axis, the angular acceleration is related to the translational acceleration by,

$a_t=r\mathrm{Î\pm }.......\left(3\right)$

If a particle moves in a straight line with a constant acceleration , its motion is described by the kinematics equations:

$$\begin{gathered} {v^2}_{xf}={v^2}_{xi}+2a_x∆x........\left(4\right) \\ ∆x=v_{xi}t+\frac{1}{2}a{t}^2......\left(5\right) \end{gathered}$$

The figure is shown below for reference.

Let the downward direction be positive. We consider a model the bucket as a particle under a net force in the vertical direction and using equation to it, we get,

$\begin{array}{r}∑F=mg−T=ma\\ ⇒T=m(g−a)⋯⋯\left(6\right)\end{array}$

Let the counter clockwise torques be positive.

We consider the model of a cylinder as a rigid object under a net torque and using equation to it, we get,

$\underset{}{∑}\mathrm{Ï„}=Tr=l\mathrm{Î\pm }$

where, moment of inertia of the cylinder is$I=\frac{1}{2}M{r}^2$.

Thus, using also, we get,

$\begin{array}{r}\mathrm{Tr}=\frac{1}{2}M{r}^2\frac{a}{r}\\ ⇒T=\frac{Ma}{2}⋯⋯\end{array}$

Now, from (6) and (7) ,

$\begin{array}{r}m(g−a)=\frac{Ma}{2}\\ ⇒mg−ma=\frac{Ma}{2}\\ ⇒mg=\frac{Ma}{2}+ma\\ ⇒mg=a\left(\frac{M}{2}+m\right)\\ ⇒a=\frac{mg}{\left(\frac{M}{2}+m\right)}\end{array}$

Substituting the values, we get,

$\begin{array}{r}a=\frac{\left(15.0\right)\left(9.80\right)}{\frac{1}{2}\left(12\right)+\left(15\right)}\\ a=7.00\mathrm{m}/{\mathrm{s}}^2\end{array}$

Using this value in (7),

$\begin{array}{r}T=\frac{\left(12.0\right)\left(7.00\right)}{2}\\ T=42.0\mathrm{N}\end{array}$

Hence, the tension in the rope is, T= 42.0 N .

Let us model the bucket as a particle under constant acceleration and using it to the equation (4), we get,

$\begin{array}{r}{v}_{yf}^2=v_{yi}^2+2a\mathrm{Δ}y\\ ⇒v_{yf}^2=2a\mathrm{Δ}y\left(âˆ\mu v_{yi}=0\right)\\ ⇒v_{yf}=\sqrt{2a\mathrm{Δ}y}\\ ⇒v_{yf}=\sqrt{2\left(7.00\right)\left(10.0\right)}\\ ⇒v_{yf}=11.8\mathrm{m}/\mathrm{s}\end{array}$

Hence, the speed of the bucket is,${\mathrm{v}}_{\mathrm{yf}}=11.8\mathrm{m}/\mathrm{s}$ .

The time of the fall of the bucket is calculated using equation to the bucket between the initial and final states:

$\begin{array}{r}\mathrm{Δ}y=v_{xi}\mathrm{Δ}t+\frac{1}{2}a(\mathrm{Δ}t{)}^2\\ ⇒\mathrm{Δ}y=\frac{1}{2}a(\mathrm{Δ}t{)}^2\\ ⇒\mathrm{Δ}t=\sqrt{\frac{2\left(\mathrm{Δ}y\right)}{a}}\\ ⇒\mathrm{Δ}t=\sqrt{\frac{2\left(10.0\right)}{7}}\\ ⇒\mathrm{Δ}t=1.69\mathrm{s}\end{array}$

Hence, the time of fall is,$∆t=1.69s$ .

The force exerted by the axle on the bucket must balance the gravitational weight of the bucket and the tension in the rope.

Thus,

$\begin{array}{r}n=T+mg\\ ⇒n=42.0+\left(12.0\right)\left(9.80\right)\\ ⇒n=160\mathrm{N}\end{array}$

Hence, the force exerted on the cylinder is, n = 160 N .