A uniform disc, of mass 2 kg and radius 0.2 m, is free to rotate in a vertical plane about a smooth horizontal axis through its centre. One end of a light inextensible string is attached to a point on the rim of the disc and the string is wound round the rim. The other end of the string is attached to a small block of mass 4 kg, which hangs freely (see diagram). The system is released from rest. During the subsequent motion, the block experiences a constant resistance to its motion, of magnitude R N. Given that the angular speed of the disc after it has turned through 2 radians is 5 rad s⁻¹, find R and the tension in the string.

A-Level CIE Further Maths 14.1 Work, Energy & Power: A uniform disc, of mass 2 kg and

A uniform disc, of mass 2 kg and radius 0.2 m, is free to rotate in a vertical plane about a smooth horizontal axis through its centre - CIE - A-Level Further Maths - Question 3 - 2013 - Paper 1

Question 3

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A uniform disc, of mass 2 kg and radius 0.2 m, is free to rotate in a vertical plane about a smooth horizontal axis through its centre. One end of a light inextensib... show full transcript

Worked Solution & Example Answer:A uniform disc, of mass 2 kg and radius 0.2 m, is free to rotate in a vertical plane about a smooth horizontal axis through its centre - CIE - A-Level Further Maths - Question 3 - 2013 - Paper 1

Step 1

Find the equation of motion for the block:

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Answer

For the block of mass 4 kg, the forces acting on it are its weight (4g) downward and the tension (T) upward. Hence, we can write the equation as:

$4g - T - R = 4a$

where 'a' is the acceleration of the block.

Step 2

Find the equation for the disc:

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Answer

For the disc, we can use the relation between angular acceleration and linear acceleration. Since the tension in the string also contributes to the rotational motion of the disc:

$T imes r = I imes \alpha$

Here, with radius r = 0.2 m, the moment of inertia I of the disc is:

$I = \frac{1}{2} m r^2 = \frac{1}{2} imes 2 imes (0.2)^2 = 0.04$

Thus, we have:

$T imes 0.2 = 0.04 \times \alpha$

From the given angular speed and distance:

$\omega^2 = 2\alpha \theta \implies 5^2 = 2\alpha imes 2$

This gives:

$\alpha = \frac{25}{4} = 6.25 \, \text{rad/s}^2$

Step 3

Substitute to find T:

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Answer

Substituting (\alpha) into the equation for the disc, we can find T:

T = \frac{0.04 \times 6.25}{0.2} = 1.25 \, \text{N}$$

Step 4

Substitute to find R:

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Now we can substitute back to find R in the equation for the block:

4(9.81) - 1.25 - R = 4(1.25)\ R = 39.24 - 1.25 - 5 = 32.99 \, \text{N}$$

A uniform disc, of mass 2 kg and radius 0.2 m, is free to rotate in a vertical plane about a smooth horizontal axis through its centre - CIE - A-Level Further Maths - Question 3 - 2013 - Paper 1

Worked Solution & Example Answer:A uniform disc, of mass 2 kg and radius 0.2 m, is free to rotate in a vertical plane about a smooth horizontal axis through its centre - CIE - A-Level Further Maths - Question 3 - 2013 - Paper 1

Find the equation of motion for the block:

Find the equation for the disc:

Substitute to find T:

Substitute to find R:

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