Question

77. Review. As shown in Figure ,$\mathbf{P}\mathbf{10}\mathbf{.}\mathbf{77}$two blocks are connected by a string of negligible mass passing over a pulley of radius $\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}\mathit{m}$ and moment ofinertia$\mathit{I}$.The block on the frictionless incline is moving with a constant acceleration of magnitude$\mathit{a}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. From this information, we wish to find the moment of inertia of the pulley. (a) What analysis model is appropriate for the blocks? (b) What analysis model is appropriate for the pulley? (c) From the analysis model in part (a) find thetension.${\mathit{T}}_{\mathbf{1}}$(d) Similarly, find the tensionlocalid="1663650764541" ${\mathit{T}}_{\mathbf{2}}$ . (e) From the analysis model in part (b), find a symbolic expression for the moment of inertia of the pulley in terms of the tensions${\mathit{T}}_{\mathbf{1}}$ and ${\mathit{T}}_{\mathbf{2}}$, the pulleyradius ,and the acceleration$\mathit{r}$ . (f) Find the numerical value of the moment of inertia of the pulley.

Short answer

(a) Analysis model appropriate for the blocks is, particle under a net force

(b) Analysis model appropriate for the pulley is, rigid object under a net torque

c) The tension$T_1$is 118 N

(d) The tension$T_2$is 156 N

(e) The moment of inertia of the pulley in terms of the tensions,$I=\frac{(T_2-T_1)r^2}{a}$

(f) The numerical value of the moment of inertia of the pulley is $1.17\text{kg}\cdot {\text{m}}^2$

Step-by-step solution

of 7: Newton’s second law of motion

Newton's second law quantifies how a force affects a body's motion. It asserts that a body's momentum change rate is proportional to the force acting on it.

${\mathit{F}}_{\mathbf{\text{net}}}\mathbf{=}\mathit{m}\mathit{a}$

Where,$\mathit{m}$ is the mass of the particle or object and$\mathit{a}$ is acceleration.

of 7: Analysis model appropriate for the blocks

(a)

In the situation of blocks provided, the appropriate model of the block is the block which has uniform mass distribution, the centre of the block which is mass concentrated and also the string attached to that point.

of 7: Analysis model appropriate for the pulley

(b)

From the provided situation of block, let us consider that the pulley is rigid.

It is also known that slipping does not occur between pulley and the string passes over it.

of 7: Calculating the tension,T1 

(c)

Given below is the free body diagram of the block of mass.$m_1$

Let us apply the Newton’s second law of motion to this free body diagram.

The amount of force acting on the mass$m_1$is as follows:

$T_1-m_{1}g\sin {37.0}^{°}=m_{1}a$

Let us substitute $15.0\text{kg}$for $m_1,$$9.8\text{m}/{\text{s}}^{2}g$and $2.00\text{m}/{\text{s}}^2$for in the above expression and solve

$\begin{array}{c}{T}_1=(15.0\text{kg})((9.80\text{m}/{\text{s}}^2)\left(\sin {37.0}^{°}\right)+(2.00\text{m}/{\text{s}}^2))\\ =118.5\text{N}\end{array}$

Therefore$T_1$,in the string is

of 7: Calculating the tension,   T2

(d)

Here comes the free diagram of the mass$m_1$

Let us apply the Newton’s second law of motion to this free body diagram.

The amount of force acting on the mass $m_1$is as follows:

$m_{2}g-T_2=m_{2}a$

Let us substitute$15.0\text{kg}$for$m_1,$$9.8\text{m}/{\text{s}}^{2}g$and $2.00\text{m}/{\text{s}}^2$forin the above expression and solve

$\begin{array}{l}(20.0\text{kg})(9.8\text{m}/{\text{s}}^2)-T_2=(20.0\text{kg})(2.00\text{m}/{\text{s}}^2)\\ T_2=(20.0\text{kg})(9.8\text{m}/{\text{s}}^2)-(20.0\text{kg})(2.00\text{m}/{\text{s}}^2)\\ T_2=156\text{N}\end{array}$

Therefore,$T_2$in the string is $156\text{N}$

of 7: Moment of inertia of the pulley in terms of the tensions  T1 and T2 ,

(e)

The torque which resulted due to the rotational inertia of the pulley is same as the resultant torque on the pulley.

$(T_2-T_1)r=I\alpha$, where$I$ is the inertia of the pulley,$\alpha$is the angular acceleration and$r$is the radius of the pulley.

The angular acceleration in terms of linear acceleration is expressed as $\alpha =\frac{a}{r}$

Let us substitute this in the torque equation and solve for inertia. We get

$\begin{array}{c}(T_2-T_1)r=I\left(\frac{a}{r}\right)\\ I=\frac{(T_2-T_1)r^2}{a}\end{array}$

Therefore, the expression of moment of inertia of pulley is $I=\frac{(T_2-T_1)r^2}{a}$

of 7: Numerical value of the moment of inertia of the pulley

(f)

From the above substitution, we get the amount of inertia of pulley as .

$I=\frac{(T_2-T_1)r^2}{a}$

Substituting$156\text{N}$, for$T_2$, $118.5\text{N}$ for $T_1$, $0.250\text{m}$for $r$,and $2.00\text{m}/{\text{s}}^2$ for in the above expression and solve, we get

$\begin{array}{c}I=\frac{(156\text{N}-118.5\text{N}){(0.250\text{m})}^2}{2.00\text{m}/{\text{s}}^2}\\ =1.17\text{kg}\cdot {\text{m}}^2\end{array}$

Hence, the value of the inertia of the pulley is $1.17\text{kg}\cdot {\text{m}}^2$