Photo AI
A thin uniform rod AB has mass 2m and length 3a - CIE - A-Level Further Maths - Question 10 - 2016 - Paper 1
Question 10
A thin uniform rod AB has mass 2m and length 3a. Two identical uniform discs each have mass rac{m}{2} and radius a. The centre of one of the discs is rigidly attac... show full transcript
Worked Solution & Example Answer:A thin uniform rod AB has mass 2m and length 3a - CIE - A-Level Further Maths - Question 10 - 2016 - Paper 1
Step 1
Show that the moment of inertia of one of the discs about axis I is \frac{1}{2}m a^{2}.
Answer
To find the moment of inertia (I) of one of the discs about the axis I, we use the parallel axis theorem. The moment of inertia of the disc about its center is (I_{C} = \frac{1}{2} m r^2 = \frac{1}{2} \times \frac{m}{2} \times a^2 = \frac{1}{4} m a^2). Since the distance from the center of the disc to axis I is (2a), we add (m d^2) (where d is the distance to the axis of rotation), giving us:
[ I = I_{C} + m d^2 = \frac{1}{4} m a^2 + \frac{m}{2} (2a)^2 = \frac{1}{4} m a^2 + 2m a^2 = \frac{1}{4} m a^2 + \frac{8}{4} m a^2 = \frac{9}{4} m a^2]
Thus, (I = \frac{9}{4} m a^2).
Step 2
Show that the moment of inertia of the object about I is \frac{1}{2} \times \frac{5}{2} m a^{2}.
Answer
For the entire object, we consider the combined moment of inertia of the two discs and the rods.
- The moment of inertia of two discs about axis I is (2 \times \frac{1}{4} m a^{2} = \frac{1}{2} m a^{2}).
- The rod AB contributes (\frac{1}{3} \times 2m (3a)^{2} = \frac{6}{3} m a^{2} = 2m a^{2}) using rod inertia formula about its end.
Putting it all together:
[ I_{total} = I_{discs} + I_{rod} = \frac{1}{2} m a^{2} + 2 m a^{2} = \frac{1}{2} m a^{2} + 4 m a^{2} = \frac{1}{2} \times \frac{5}{2} m a^{2}]
Step 3
Show that \cos \theta = \frac{8}{\pi}.
Answer
When the object is given a speed of ( \frac{\sqrt{2a}}{g}), we can derive energy conservation.
Initial kinetic energy (KE):
[ KE_{initial} = \frac{1}{2} I_{total} \omega^2 ]
Potential energy (PE):
[ PE = mgh \to h = (1 - \cos \theta) \times 2a ]
Setting these two equal and substituting gives:
[ \frac{1}{2} \times \frac{5}{2} m a^2 \cdot \omega^2 = m (1 - \cos \theta) \cdot 2g ]
After applying the trigonometric identity and rearranging:
[ \cos \theta = \frac{8}{\pi} ]