I am struggling to solve the problem. Would love some assisstance on it as the homework is due in a couple hours. Two blocks, m=3 kg and M=5.8 kg, are hanging on an Atwood's machine. The mass ofthe pulley is 10 kg and the radius is 0.15 m. We can assume this is a solid disk and ignorefriction.What is the moment of inertia of the pulley?The blocks are initially held at rest, then released. At the instant the larger mass M hasdropped 1.5 m, what is the speed of each block? Please remember to draw free-bodydiagrams.What is the tension in the rope at that same instant?

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Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

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I am struggling to solve the problem. Would love some assisstance on it as the homework is due in a couple hours.

Two blocks, m=3 kg and M=5.8 kg, are hanging on an Atwood's machine. The mass of
the pulley is 10 kg and the radius is 0.15 m. We can assume this is a solid disk and ignore
friction.
What is the moment of inertia of the pulley?
The blocks are initially held at rest, then released. At the instant the larger mass M has
dropped 1.5 m, what is the speed of each block? Please remember to draw free-body
diagrams.
What is the tension in the rope at that same instant?

Expert Solution

Step 1 :Introduction

The free body diagram of the system is :

Due to the moment of inertia of the pulley , the tension $T_{2}$ and $T_{1}$ will not be the same . The change in the tension on either side of the pulley , will provide an torque to the pulley , which is calculated by using the formula , $τ = I α$ , where $I = \frac{1}{2} M_{p} r^{2}$ is the moment of inertia of a solid disk , $M_{p} = 10 k g$ is the mass of pulley , $r = 0.15 m$ is the radius of the pulley , $α = \frac{a}{r}$ is the angular acceleration of the pulley and a is the acceleration of the system . Thus ,

$\begin{array}{rcl} τ & = & I α \\ T_{2} r - T_{1} r & = & \frac{1}{2} M_{p} r^{2} \times (\frac{a}{r}) \\ T_{2} - T_{1} & = & \frac{1}{2} M_{p} a \to (1) \end{array}$

According to the Newton's second law , the net force acting on block $m = 3 k g$ is $\sum_{} F = m a$ , where the forces acting are the torque $T_{1}$ upward and weight force downward , which is , $F_{m} = m g$ . Thus ,

$\begin{array}{rcl} \sum_{} F & = & m a \\ T_{1} - m g & = & m a \\ T_{1} & = & m a + m g \to (2) \end{array}$

According to the Newton's second law , the net force acting on block $M = 5.8 k g$ is $\sum_{} F = M a$ , where the forces acting are the torque $T_{2}$ upward and weight force downward , which is , $F_{M} = M g$ . Thus ,

$\begin{array}{rcl} \sum_{} F & = & M a \\ M g - T_{2} & = & M a \\ T_{2} & = & M g - M a \to (3) \end{array}$

From (2) and (3) equation (1) becomes

$\begin{array}{rcl} T_{2} - T_{1} & = & \frac{1}{2} M_{p} a \\ (M g - M a) - (m a + m g) & = & \frac{1}{2} M_{p} a \\ M g - M a - m a - m g & = & \frac{1}{2} M_{p} a \\ g (M - m) & = & \frac{1}{2} M_{p} a + M a + m a \\ = & a (\frac{1}{2} M_{p} + M + m) \\ a & = & \frac{g (M - m)}{(\frac{1}{2} M_{p} + M + m)} \to (4) \end{array}$

According to the law of conservation of energy , energy can neither be created nor destroyed , but can only be converted from one form to another . Thus the initial energy in the system which is the potential energy of the two blocks , will be converted to their final potential energy after dropping a height of $∆ h = 1.5 m$ , their kinetic energy and the rotational energy of the pulley , which is , $E_{r} = \frac{1}{2} I ω^{2}$ , where $ω = \frac{v}{r}$ is the angular velocity and v is the velocity of the system .Thus ,

$\begin{array}{rcl} M g h_{i} + m g h_{i}' & = & M g h_{f} + m g h_{f}' + \frac{1}{2} M v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} \\ M g h_{f} + m g h_{f}' - M g h_{i} - m g h_{i}' + \frac{1}{2} M v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} & = & 0 \\ - M g ∆ h + m g ∆ h + \frac{1}{2} M v^{2} + \frac{1}{2} m v^{2} + \frac{1}{2} \times (\frac{1}{2} M_{p} r^{2}) \times {(\frac{v}{r})}^{2} & = & 0 \\ - M g ∆ h + m g ∆ h + \frac{1}{2} M v^{2} + \frac{1}{2} m v^{2} + \frac{1}{4} M_{p} v^{2} & = & 0 \\ g ∆ h (- M + m) + v^{2} (\frac{1}{2} M + \frac{1}{2} m + \frac{1}{4} M_{p}) & = & 0 \\ v & = & \sqrt{\frac{g ∆ h (M - m)}{(\frac{1}{2} M + \frac{1}{2} m + \frac{1}{4} M_{p})}} \to (5) \end{array}$

, where $h_{i}$ is the initial height of block $M = 5.8 k g$ , $h_{i}'$ is the initial height of block $m = 3 k g$ , $h_{f}$ is the final height of block $M = 5.8 k g$ , $h_{f}'$ is the final height of block $m = 3 k g$ and v is the final velocity of the blocks.

Step 2 :Calculation of moment of inertia

Calculate moment of inertia of the pulley by using the formula $I = \frac{1}{2} M_{p} r^{2}$ , where $M_{p} = 10 k g$ is the mass of pulley and $r = 0.15 m$ is the radius of the pulley .

$\begin{array}{rcl} I & = & \frac{1}{2} M_{p} r^{2} \\ = & \frac{1}{2} \times (10 k g) \times {(0.15 m)}^{2} \\ = & 0.1125 k g m^{2} \end{array}$

Step 3 :Calculation of the speed of the blocks

Calculate the speed of the blocks by using relation (5) , which is , $\begin{array}{rcl} v & = & \sqrt{\frac{g ∆ h (M - m)}{(\frac{1}{2} M + \frac{1}{2} m + \frac{1}{4} M_{p})}} \to (5) \end{array}$ , where $g = 9.8 m / s^{2}$ is the acceleration due to gravity , $∆ h = 1.5 m$ is the change in height , $M_{p} = 10 k g$ is the mass of pulley , $M = 5.8 k g$ and $m = 3 k g$ .

$\begin{array}{rcl} v & = & \sqrt{\frac{g ∆ h (M - m)}{(\frac{1}{2} M + \frac{1}{2} m + \frac{1}{4} M_{p})}} \\ = & \sqrt{\frac{(9.8 m / s^{2}) \times (1.5 m) (5.8 k g - 3 k g)}{(\frac{1}{2} (5.8 k g) + \frac{1}{2} (3 k g) + \frac{1}{4} (10 k g))}} \\ \approx & 2.44 m / s \end{array}$

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