In the figure, two 6.70 kg blocks are connected by a massless string over a pulley of radius 2.20 cm and rotational inertia
. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block’the
pulley’s axis is frictionless. When this system is released from rest, the pulley turns through 1.20 rad in 102 ms and the acceleration of
the blocks is constant. What are (a) the magnitude of the pulley’s angular acceleration, (b) the magnitude of either block’s acceleration,
© string tension , and (d) string tension ? Assume free-fall acceleration to be equal to .
(a) Number 230
(b) Number 5.06
© Number 31.8
(d) Number
(i Units

Ask by Boyd Wood. in the United States

Mar 27,2025

Question

In the figure, two 6.70 kg blocks are connected by a massless string over a pulley of radius 2.20 cm and rotational inertia
. The string does not slip on the pulley; it is not known whether there is friction between the table and the sliding block’the
pulley’s axis is frictionless. When this system is released from rest, the pulley turns through 1.20 rad in 102 ms and the acceleration of
the blocks is constant. What are (a) the magnitude of the pulley’s angular acceleration, (b) the magnitude of either block’s acceleration,
© string tension , and (d) string tension ? Assume free-fall acceleration to be equal to .
(a) Number 230
(b) Number 5.06
© Number 31.8
(d) Number
(i Units

Ask by Boyd Wood. in the United States

Mar 27,2025

UpStudy AI · Accepted Answer

To solve this problem, we will follow these steps:

1. **Calculate the angular acceleration of the pulley**.
2. **Determine the linear acceleration of the blocks**.
3. **Calculate the tensions in the string**.

### Given Data:
- Mass of each block, $$ m = 6.70 \, \text{kg} $$
- Radius of the pulley, $$ r = 2.20 \, \text{cm} = 0.022 \, \text{m} $$
- Rotational inertia of the pulley, $$ I = 7.40 \times 10^{-4} \, \text{kg} \cdot \text{m}^{2} $$
- Angle turned by the pulley, $$ \theta = 1.20 \, \text{rad} $$
- Time taken, $$ t = 102 \, \text{ms} = 0.102 \, \text{s} $$
- Acceleration due to gravity, $$ g = 9.81 \, \text{m/s}^{2} $$

### Step 1: Calculate the angular acceleration of the pulley
The angular displacement $$ \theta $$ is related to angular acceleration $$ \alpha $$ and time $$ t $$ by the equation:
$$
\theta = \frac{1}{2} \alpha t^2
$$
Rearranging for $$ \alpha $$:
$$
\alpha = \frac{2\theta}{t^2}
$$

Now, substituting the values:
$$
\alpha = \frac{2 \times 1.20}{(0.102)^2}
$$

### Step 2: Calculate the linear acceleration of the blocks
The linear acceleration $$ a $$ of the blocks is related to the angular acceleration $$ \alpha $$ by:
$$
a = r \alpha
$$

### Step 3: Calculate the tensions in the string
Using Newton's second law for one of the blocks, we can write:
$$
m g - T_1 = m a \quad \text{(for the block going down)}
$$
$$
T_2 - m g = m a \quad \text{(for the block going up)}
$$

The net torque on the pulley is given by:
$$
\tau = I \alpha = (T_1 - T_2) r
$$

Now we can solve for $$ T_1 $$ and $$ T_2 $$.

Let's perform the calculations for $$ \alpha $$, $$ a $$, $$ T_1 $$, and $$ T_2 $$.

First, we will calculate $$ \alpha $$ and $$ a $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{2\times 1.2}{0.102^{2}}$$
- step1: Convert the expressions:
  $$\frac{2\times 1.2}{\left(\frac{51}{500}\right)^{2}}$$
- step2: Multiply the numbers:
  $$\frac{2.4}{\left(\frac{51}{500}\right)^{2}}$$
- step3: Convert the expressions:
  $$\frac{\frac{12}{5}}{\left(\frac{51}{500}\right)^{2}}$$
- step4: Evaluate the power:
  $$\frac{\frac{12}{5}}{\frac{2601}{500^{2}}}$$
- step5: Multiply by the reciprocal:
  $$\frac{12}{5}\times \frac{500^{2}}{2601}$$
- step6: Reduce the numbers:
  $$\frac{4}{5}\times \frac{500^{2}}{867}$$
- step7: Rewrite the expression:
  $$\frac{4}{5}\times \frac{125^{2}\times 4^{2}}{867}$$
- step8: Rewrite the expression:
  $$\frac{4}{5}\times \frac{5^{6}\times 4^{2}}{867}$$
- step9: Reduce the numbers:
  $$4\times \frac{5^{5}\times 4^{2}}{867}$$
- step10: Multiply:
  $$\frac{4\times 50000}{867}$$
- step11: Multiply:
  $$\frac{200000}{867}$$
Calculate or simplify the expression $$ 0.022*(2*1.20/(0.102^2)) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$0.022\left(\frac{2\times 1.2}{0.102^{2}}\right)$$
- step1: Convert the expressions:
  $$0.022\left(\frac{2\times 1.2}{\left(\frac{51}{500}\right)^{2}}\right)$$
- step2: Multiply the numbers:
  $$0.022\left(\frac{2.4}{\left(\frac{51}{500}\right)^{2}}\right)$$
- step3: Divide the terms:
  $$0.022\times \frac{200000}{867}$$
- step4: Convert the expressions:
  $$\frac{11}{500}\times \frac{200000}{867}$$
- step5: Multiply the fractions:
  $$\frac{11\times 200000}{500\times 867}$$
- step6: Multiply the terms:
  $$\frac{2200000}{500\times 867}$$
- step7: Multiply the terms:
  $$\frac{2200000}{433500}$$
- step8: Reduce the fraction:
  $$\frac{4400}{867}$$
### Step 1: Calculate the angular acceleration $$ \alpha $$
The angular acceleration is calculated as:
$$
\alpha = \frac{2 \times 1.20}{(0.102)^2} \approx 230 \, \text{rad/s}^2
$$

### Step 2: Calculate the linear acceleration $$ a $$
The linear acceleration is given by:
$$
a = r \alpha = 0.022 \times 230 \approx 5.06 \, \text{m/s}^2
$$

### Step 3: Calculate the tensions $$ T_1 $$ and $$ T_2 $$
Using the equations of motion for the blocks and the torque equation for the pulley:

1. For the block going down:
   $$
   m g - T_1 = m a
   $$
   Rearranging gives:
   $$
   T_1 = m g - m a
   $$

2. For the block going up:
   $$
   T_2 - m g = m a
   $$
   Rearranging gives:
   $$
   T_2 = m g + m a
   $$

3. The net torque on the pulley:
   $$
   I \alpha = (T_1 - T_2) r
   $$

Now, substituting the values to find $$ T_1 $$ and $$ T_2 $$:
- $$ m = 6.70 \, \text{kg} $$
- $$ g = 9.81 \, \text{m/s}^2 $$
- $$ a = 5.06 \, \text{m/s}^2 $$

Let's calculate $$ T_1 $$ and $$ T_2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$6.7\times 9.81+6.7\times 5.06$$
- step1: Multiply the numbers:
  $$65.727+6.7\times 5.06$$
- step2: Multiply the numbers:
  $$65.727+33.902$$
- step3: Add the numbers:
  $$99.629$$
Calculate or simplify the expression $$ 6.70*9.81 - 6.70*5.06 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$6.7\times 9.81-6.7\times 5.06$$
- step1: Multiply the numbers:
  $$65.727-6.7\times 5.06$$
- step2: Multiply the numbers:
  $$65.727-33.902$$
- step3: Subtract the numbers:
  $$31.825$$
### Step 3: Calculate the tensions $$ T_1 $$ and $$ T_2 $$

1. **Tension $$ T_1 $$**:
   $$
   T_1 = 6.70 \times 9.81 - 6.70 \times 5.06 \approx 31.83 \, \text{N}
   $$

2. **Tension $$ T_2 $$**:
   $$
   T_2 = 6.70 \times 9.81 + 6.70 \times 5.06 \approx 99.63 \, \text{N}
   $$

### Summary of Results:
- (a) The magnitude of the pulley's angular acceleration $$ \alpha $$ is approximately $$ 230 \, \text{rad/s}^2 $$.
- (b) The magnitude of either block's acceleration $$ a $$ is approximately $$ 5.06 \, \text{m/s}^2 $$.
- (c) The tension $$ T_1 $$ is approximately $$ 31.83 \, \text{N} $$.
- (d) The tension $$ T_2 $$ is approximately $$ 99.63 \, \text{N} $$.
(a) The pulley's angular acceleration is 230 rad/s². (b) The acceleration of the blocks is 5.06 m/s². (c) The tension $$ T_1 $$ is 31.8 N. (d) The tension $$ T_2 $$ is 99.6 N.

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