91影视

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鈥檚 Second Law:

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

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,

$\underset{}{鈭�}{蟿}_{ext}=l伪.......\left(2\right)$

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

$a_t=r伪.......\left(3\right)$

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

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

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,

$\begin{array}{r}鈭�F=mg鈭�T=ma\\ 鈬�T=m(g鈭�a)鈰�鈰�\left(6\right)\end{array}$

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,

$\underset{}{鈭�}蟿=Tr=l伪$

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

Thus, using also, we get,

$\begin{array}{r}\mathrm{Tr}=\frac{1}{2}M{r}^2\frac{a}{r}\\ 鈬�T=\frac{Ma}{2}鈰�鈰�\end{array}$

Now, from (6) and (7) ,

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

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,

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

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:

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

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

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

Thus,

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

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