Question

I'm having a really hard time trying to solve this

A block with mass mb = 1.3 kg is connected by a rope across a 50-cm-diameter, 2.0 kg pulley, as shown in (Figure 1). There is no friction in the axle, but there is friction between the rope and the pulley; the rope doesn't slip and the pulley can be modeled as a solid cylinder. The weight is accelerating upward at 1.2 m/s^2.

What is the tension in the rope on the right side of the pulley?

Answer 1

Approximately
`15.5\; {\rm N}`, assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.

Step-by-step explanation:

To find the tension in the rope on the right side of the pulley, apply the following steps:

• Find the tension that the rope exerts on the block, which is equal to the tension
  `T_{\text{left}}` on the left side of the pulley.
• Find the torque
  `\tau_{\text{left}}` resulting from the tension
  `T_{\text{left}}` on the left side of the pulley.
• Find the moment of inertia
  `I` of the pulley and the net torque
  `\tau_{\text{net}}`.
• Add the torque on the left
  `\tau_{\text{left}}` to the net torque
  `\tau_{\text{net}}` to find
  `\tau_{\text{right}}`, the torque on the right side of the pulley.
• Divide
  `\tau_{\text{right}}` by radius of the pulley
  `r` to find the tension on the right side,
  `T_{\text{right}}`.

The net force on the block is:

`F_{\text{net}} = m_{\text{b}} \, a`, where

• 
  `m_{\text{b}} = 1.3\; {\rm kg}` is the mass of the block, and
• 
  `a = 1.2\; {\rm m\cdot s^(-2)}` is the linear acceleration of the block.

At the same time, the net force on the block can also be expressed as:

`\begin{aligned}F_{\text{net}} &= T_{\text{left}} - (\text{weight}) \\ &= T_{\text{left}} - m_{\text{b}}\, g \end{aligned}`, where

• 
  `g = 9.81\; {\rm N\cdot kg^(-1)}` by assumption, and
• 
  `T_{\text{left}}` is the tension the rope exerted on the block. This tension is equal to the tension on the left side of the pulley.

Rearrange and solve for
`T_{\text{left}}`:

`T_{\text{left}} - m_{\text{b}}\, g = F_{\text{net}} = m_{\text{b}}\, a`.

`\begin{aligned}T_{\text{left}} &= m_{\text{b}}\, a + m_{\text{b}}\, g \\ &= m_{\text{b}}\, (a + g) \\ &= 1.3\, (1.2 + 9.81)\; {\rm N} \\ &= 14.313\; {\rm N}\end{aligned}`.

Let
`r` denote the radius of the pulley. It is given that the diameter of the pulley is
`50\; {\rm cm}`. In standard units, the radius of the pulley would be
`r = 25\; {\rm cm} = 0.25\; {\rm m}`.

On the left side of the pulley, tension in the rope exerts a torque of
`\tau_{\text{left}} = T_{\text{left}}\, r` on the pulley:

`\begin{aligned}\tau_{\text{left}} &= T_{\text{left}}\, r \\ &= (14.313)\, (0.25)\; {\rm N\cdot m} \\ &= 3.57825\; {\rm N\cdot m} \end{aligned}`.

Under the assumptions, the moment of inertia
`I` of this cylindrical pulley would be:

`\begin{aligned} I &= (1)/(2)\, m\, r^(2) \end{aligned}`, where

• 
  `m = 2.0\; {\rm kg}` is the mass of the pulley, and
• 
  `r = 0.25\; {\rm m}` is the radius of the pulley.

`\begin{aligned} I &= (1)/(2)\, m\, r^(2) \\ &= (1)/(2)\, (2.0)\, (0.25)^(2)\; {\rm kg \cdot m^(2)} \\ &= 0.0625\; {\rm kg\cdot m^(2)} \end{aligned}`.

Since the rope doesn't slip on the pulley, linear acceleration of the pulley would be equal to that of the rope,
`a = 1.2\; {\rm m\cdot s^(-2)}`. Divide this linear acceleration by the radius of the pulley to find the angular acceleration
`\alpha` of the pulley:

`\begin{aligned}\alpha &= (a)/(r) \\ &= (1.2)/(0.25)\; {\rm s^(-2)} \\ &= 4.8\; {\rm s^(-2)}\end{aligned}`.

Multiply angular acceleration by the moment of inertia to find the net torque
`\tau_{\text{net}}` on the pulley cylinder:

`\begin{aligned}\tau_{\text{net}} &= I\, \alpha \\ &= (0.0625)\, (4.8)\; {\rm kg \cdot m^(2)\cdot s^(-2)}\\ &= 0.3\; {\rm kg \cdot m^(2) \cdot s^(-2)} \end{aligned}`.

Note that the net torque of the pulley
`\tau_{\text{net}}` is in the same direction as
`\tau_{\text{right}}`, but the opposite of
`\tau_{\text{left}}`. Hence:

`\begin{aligned}\tau_{\text{right}} &= \tau_{\text{net}} + \tau_{\text{left}} \\ &= 0.3\; {\rm N\cdot m} + 3.57825\; {\rm N\cdot m} \\ &= 3.87825\; {\rm N\cdot m}\end{aligned}`.

Divide the torque on the right
`\tau_{\text{right}}` by radius
`r` to find the tension in the string on the right
`T_{\text{right}}`:

`\begin{aligned}T_{\text{right}} &= \frac{\tau_{\text{right}}}{r} \\ &= (3.87825)/(0.25)\; {\rm N} \\ &= 15.513\; {\rm N}\end{aligned}`.