91影视

Given in the question,

The radius of disk 1 is,$R_1=2.50\text{鈥夆赌塩m}$

The mass of disk 1 is,$M_1=0.80\text{鈥夆赌塳g}$

The radius of disk 2 is,$R_1=5.00\text{鈥夆赌塩m}$

The mass of disk 1 is,$M_2=1.60\text{鈥夆赌塳g}$

The mass of the block$m=1.50\text{鈥夆赌塳g}$

Distance covered by the block $d=2\text{鈥夆赌塵}$

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鈥檚 third law, for rotational motion

$I伪=蟿$

Where I is inertia , $蟿$ is torques, and $伪$is angular acceleration.

4. Relation between linear and angular acceleration

$a=r伪$

Where a is linear acceleration, r is radius and $伪$ is the angular acceleration

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

$\begin{array}{l}{I}_{total}=I_1+I_2\\ I_{total}=\frac{1}{2}{M}_1{R}_1^2+\frac{1}{2}{M}_2{R}_2^2\\ I_{total}=\frac{1}{2}\left(0.80\text{鈥夆赌塳g}\right){\left(0.025\text{鈥夆赌塵}\right)}^2+\frac{1}{2}\left(1.60\text{鈥夆赌塳g}\right){\left(0.05\text{鈥夆赌塵}\right)}^2\\ I_{total}=0.00025{\text{鈥夆赌塳g.m}}^{\text{2}}+0.002{\text{鈥夆赌塳g.m}}^{\text{2}}\\ I_{total}=0.00225{\text{鈥夆赌塳g.m}}^{\text{2}}\end{array}$

Hence the total moment of inertia of the system is$0.00225{\text{鈥夆赌塳g.m}}^{\text{2}}$

Let the tension in the string is T.

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

$\begin{array}{l}{F}_{net}=ma\\ mg鈭�T=ma\text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌夆��}\left(i\right)\\ I_{total}伪=蟿\end{array}$

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

Therefore, we can write

$\begin{array}{l}{I}_{total}伪=T.R_1\\ T=\frac{I_{total}伪}{R_1}\text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌夆�夆赌�}\left(ii\right)\end{array}$

Now substituting this value into equation (i)

$mg鈭�\left(\frac{I_{total}伪}{R_1}\right)=ma$

We know,

Linear acceleration = angular acceleration multiplied by the radius

$\begin{array}{l}a=R_1伪\\ 伪=\frac{a}{R_1}\end{array}$

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

Substituting the values

$\begin{array}{l}a=\frac{\left(1.50\text{鈥夆赌塳g}\right)\left(9.8\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\right)}{\left(\left(1.50\text{鈥夆赌塳g}\right)+\frac{0.00225{\text{鈥夆赌塳g.m}}^{\text{2}}}{{\left(0.0250\text{鈥夆赌塵}\right)}^2}\right)}\\ a=2.88\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\end{array}$

Acceleration of the block is $a=2.88\text{鈥夆赌塵}/{\text{s}}^{\text{2}}$

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

$\begin{array}{l}{v}^2=0^2+2\left(2.88\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\right)\left(2\text{鈥夆赌塵}\right)\\ v=3.4\text{m}/\text{s}\end{array}$

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

$\begin{array}{l}{v}^2=0^2+2\left(6.12\text{鈥夆赌夆�夆赌塵}/{\text{s}}^{\text{2}}\right)\left(2\text{鈥夆赌塵}\right)\\ v=4.9\text{鈥尘}/\text{s}\end{array}$

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

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鈥檚 third law for both the transitional and rotational motion.

$\begin{array}{c}{F}_{net}=ma\\ mg鈭�T=ma\text{鈥夆赌夆�夆赌夆��}\left(i\right)\\ I_{total}伪=蟿\end{array}$

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

Therefore, we can write

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

Now substituting this value into equation (i)

$mg鈭�\left(\frac{I_{total}伪}{R_2}\right)=ma$

We know

Linear acceleration = angular acceleration multiplied by the radius

$$\begin{gathered} a=R_1伪 \\ 伪=\frac{a}{R_2} \end{gathered}$$

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

Substituting the values

$\begin{array}{l}a=\frac{\left(1.50\text{鈥夆赌塳g}\right)\left(9.8\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\right)}{\left(\left(1.50\text{鈥夆赌塳g}\right)+\frac{0.00225{\text{鈥夆赌塳g.m}}^{\text{2}}}{{\left(0.050\text{鈥夆赌塵}\right)}^2}\right)}\\ a=6.12\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\end{array}$

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.