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A bob of mass 0.50 kg is suspended from the end of a piece of string 0.45 m long - AQA - A-Level Physics - Question 30 - 2017 - Paper 1

Question 30

A bob of mass 0.50 kg is suspended from the end of a piece of string 0.45 m long. The bob is rotated in a vertical circle at a constant rate of 120 revolutions per m... show full transcript

Worked Solution & Example Answer:A bob of mass 0.50 kg is suspended from the end of a piece of string 0.45 m long - AQA - A-Level Physics - Question 30 - 2017 - Paper 1

Step 1

Calculate the centripetal acceleration

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Answer

First, we need to calculate the linear velocity of the bob. Given that the bob completes 120 revolutions per minute, we can convert this to radians per second:

Convert revolutions per minute (rpm) to radians per second (rad/s):

$\text{Angular velocity} \omega = 120 \text{ rev/min} \times \frac{2\pi \ ext{rad}}{1 \ ext{rev}} \times \frac{1 \ ext{min}}{60 \ ext{s}} = 12.57 \text{ rad/s}$
The linear velocity $v$ is given by:

$v = r\omega$ , where $r = 0.45\text{ m}$ :

$v = 0.45 \text{ m} \times 12.57 \text{ rad/s} \approx 5.65 \text{ m/s}$
The centripetal acceleration $a_c$ can now be calculated:

$a_c = \frac{v^2}{r} = \frac{(5.65 \text{ m/s})^2}{0.45 \text{ m}} \approx 71.0 \text{ m/s}^2$

Step 2

Apply Newton's second law of motion

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Answer

When the bob is at the bottom of the circle, the tension $T$ in the string must support both the weight of the bob and provide the centripetal force required for circular motion:

$T - mg = ma_c$

Rearranging gives us:

$T = mg + ma_c$

Now, substituting in the known values:

Mass $m = 0.50 \text{ kg}$
Acceleration due to gravity $g = 9.81 \text{ m/s}^2$
Centripetal acceleration $a_c \approx 71.0 \text{ m/s}^2$

Substitute:

$T = (0.50 \text{ kg})(9.81 \text{ m/s}^2) + (0.50 \text{ kg})(71.0 \text{ m/s}^2)$

Calculating gives:

$T \approx 4.905 \text{ N} + 35.5 \text{ N} \approx 40.41 \text{ N}$

Step 3

Select the correct answer

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Answer

According to the calculated values, the tension in the string when the bob is at the bottom of the circle is approximately 40.41 N, which corresponds to option D: 40 N.

A bob of mass 0.50 kg is suspended from the end of a piece of string 0.45 m long - AQA - A-Level Physics - Question 30 - 2017 - Paper 1

Worked Solution & Example Answer:A bob of mass 0.50 kg is suspended from the end of a piece of string 0.45 m long - AQA - A-Level Physics - Question 30 - 2017 - Paper 1

Calculate the centripetal acceleration

Apply Newton's second law of motion

Select the correct answer

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