91Ó°ÊÓ

Given in the question,

The radius of disk 1 is,$R_1=2.50\text{ â¶Ä‰cm}$

The mass of disk 1 is,$M_1=0.80\text{ â¶Ä‰kg}$

The radius of disk 2 is,$R_1=5.00\text{ â¶Ä‰cm}$

The mass of disk 1 is,$M_2=1.60\text{ â¶Ä‰kg}$

The mass of the block$m=1.50\text{ â¶Ä‰kg}$

Distance covered by the block $d=2\text{ â¶Ä‰m}$

Rotational Motion

If the motion of an object is around a circular path, in a fixed orbit, then that motion is known as rotational motion.

Since the disk is rotating, we must apply the formula of rotational motion

1.Moment of inertia of a disk

$I=\frac{1}{2}M{R}^2$

Where I is the moment of inertia, M is mass and R is the radius.

$F=ma$

Where F is force, m is mass and a is acceleration

3. Newton’s third law, for rotational motion

$I\mathrm{Î\pm }=\mathrm{Ï„}$

Where I is inertia , $\mathrm{Ï„}$ is torques, and $\mathrm{Î\pm }$is angular acceleration.

4. Relation between linear and angular acceleration

$a=r\mathrm{Î\pm }$

Where a is linear acceleration, r is radius and $\mathrm{Î\pm }$ is the angular acceleration

The total moment of inertia will be some of the inertia of both the disks

$\begin{array}{l}{I}_{total}=I_1+I_2\\ I_{total}=\frac{1}{2}{M}_1{R}_1^2+\frac{1}{2}{M}_2{R}_2^2\\ I_{total}=\frac{1}{2}\left(0.80\text{ â¶Ä‰kg}\right){\left(0.025\text{ â¶Ä‰m}\right)}^2+\frac{1}{2}\left(1.60\text{ â¶Ä‰kg}\right){\left(0.05\text{ â¶Ä‰m}\right)}^2\\ I_{total}=0.00025{\text{ â¶Ä‰kg.m}}^{\text{2}}+0.002{\text{ â¶Ä‰kg.m}}^{\text{2}}\\ I_{total}=0.00225{\text{ â¶Ä‰kg.m}}^{\text{2}}\end{array}$

Hence the total moment of inertia of the system is$0.00225{\text{ â¶Ä‰kg.m}}^{\text{2}}$

Let the tension in the string is T.

Now applying Newton’s third law for both the transitional and rotational motion.

$\begin{array}{l}{F}_{net}=ma\\ mg−T=ma\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰}\left(i\right)\\ I_{total}\mathrm{Î\pm }=\mathrm{Ï„}\end{array}$

Considering both the disk as a system since the only force acting on the disks is T

Therefore, we can write

$\begin{array}{l}{I}_{total}\mathrm{Î\pm }=T.R_1\\ T=\frac{I_{total}\mathrm{Î\pm }}{R_1}\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰}\left(ii\right)\end{array}$

Now substituting this value into equation (i)

$mg−\left(\frac{I_{total}\mathrm{Î\pm }}{R_1}\right)=ma$

We know,

Linear acceleration = angular acceleration multiplied by the radius

$\begin{array}{l}a=R_1\mathrm{Î\pm }\\ \mathrm{Î\pm }=\frac{a}{R_1}\end{array}$

$\begin{array}{c}mg−\left(\frac{I_{total}\left(a/R_1\right)}{R_1}\right)=ma\\ a\left(m+\frac{I_{total}}{R_1^2}\right)=mg\\ a=\frac{mg}{\left(m+\frac{I_{total}}{R_1^2}\right)}\end{array}$

Substituting the values

$\begin{array}{l}a=\frac{\left(1.50\text{ â¶Ä‰kg}\right)\left(9.8\text{ â¶Ä‰m}/{\text{s}}^{\text{2}}\right)}{\left(\left(1.50\text{ â¶Ä‰kg}\right)+\frac{0.00225{\text{ â¶Ä‰kg.m}}^{\text{2}}}{{\left(0.0250\text{ â¶Ä‰m}\right)}^2}\right)}\\ a=2.88\text{ â¶Ä‰m}/{\text{s}}^{\text{2}}\end{array}$

Acceleration of the block is $a=2.88\text{ â¶Ä‰m}/{\text{s}}^{\text{2}}$

Now to find the velocity using the equation

$v^2=u^2+2ad$

Where v is final velocity, u is initial velocity, a is acceleration and d is the distance

$\begin{array}{l}{v}^2=0^2+2\left(2.88\text{ â¶Ä‰m}/{\text{s}}^{\text{2}}\right)\left(2\text{ â¶Ä‰m}\right)\\ v=3.4\text{m}/\text{s}\end{array}$

Hence the speed of the block just before it strikes the floor is $v=3.4m/s$

Where v is final velocity, u is initial velocity, a is acceleration and d is the distance

$\begin{array}{l}{v}^2=0^2+2\left(6.12\text{ â¶Ä‰â€‰â¶Ä‰m}/{\text{s}}^{\text{2}}\right)\left(2\text{ â¶Ä‰m}\right)\\ v=4.9\text{″¾}/\text{s}\end{array}$

Hence the speed of the block just before it strikes the floor is $v=4.9\text{″¾}/\text{s}$

Speed is more when the string is wrapped to the larger disk because the radius of the disk is larger, so the acceleration is more, hence the velocity of the block is more.

Now if the string is wrapped around the larger disk. Let the tension in the string is T

Now applying Newton’s third law for both the transitional and rotational motion.

$\begin{array}{c}{F}_{net}=ma\\ mg−T=ma\text{ â¶Ä‰â€‰â¶Ä‰â€‰}\left(i\right)\\ I_{total}\mathrm{Î\pm }=\mathrm{Ï„}\end{array}$

Considering both the disk as a system since the only force acting on the disks is T

Therefore, we can write

$$\begin{gathered} I_{total}\mathrm{Î\pm }=T.R_2 \\ T=\frac{I_{total}\mathrm{Î\pm }}{R_2}\left(ii\right) \end{gathered}$$

Now substituting this value into equation (i)

$mg−\left(\frac{I_{total}\mathrm{Î\pm }}{R_2}\right)=ma$

We know

Linear acceleration = angular acceleration multiplied by the radius

$$\begin{gathered} a=R_1\mathrm{Î\pm } \\ \mathrm{Î\pm }=\frac{a}{R_2} \end{gathered}$$

$\begin{array}{c}mg−\left(\frac{I_{total}\left(a/R_2\right)}{R_2}\right)=ma\\ a\left(m+\frac{I_{total}}{R_2^2}\right)=mg\\ a=\frac{mg}{\left(m+\frac{I_{total}}{R_2^2}\right)}\end{array}$

Substituting the values

$\begin{array}{l}a=\frac{\left(1.50\text{ â¶Ä‰kg}\right)\left(9.8\text{ â¶Ä‰m}/{\text{s}}^{\text{2}}\right)}{\left(\left(1.50\text{ â¶Ä‰kg}\right)+\frac{0.00225{\text{ â¶Ä‰kg.m}}^{\text{2}}}{{\left(0.050\text{ â¶Ä‰m}\right)}^2}\right)}\\ a=6.12\text{ â¶Ä‰m}/{\text{s}}^{\text{2}}\end{array}$

Acceleration of the block is $6.12m/s^2$

Now to find the velocity using the equation

$v^2=u^2+2ad$

Where v is final velocity, u is initial velocity, a is acceleration and d is the distance

\(\begin{aligned}{}{v^2} = {0^2} + 2\left( {6.12\,\,\,\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {2\,\,{\rm{m}}} \right)\\v = 4.9\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Hence the speed of the block just before it strikes the floor is \(v = 4.9\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\)

Speed is more when the string is wrapped to the larger disk because the radius of the disk is larger, so the acceleration is more, hence the velocity of the block is more.