The moment of inertial of a thin rod is:
Correct answer: D. 1/12 mL2
- A. 1/2 mL2
- B. 1/4 m3L
- C. 1/12 mL
- D. 1/12 mL2
Explanation
The moment of inertia of an object is a measure of its resistance to changes in its rotation. For a thin rod of length L and mass m, rotating about an axis through its center and perpendicular to its length, the moment of inertia is given by the formula I = (1/12) mL2. This correctly accounts for both the mass and the distribution of that mass along the length of the rod. Option A represents the moment of inertia of a different geometric shape, option B has incorrect units and a non-standard form, and option C lacks the correct dimensions for a moment of inertia. Thus, Option D is the correct and only valid choice.
This question appeared on the UHS MDCAT 2010 paper, which you can sit online with every answer explained.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Angular and Linear Quantities
Rotational motion relates angular displacement, angular velocity and angular acceleration to linear displacement, tangential velocity and tangential acceleration through the radius. Questions use radians, v equals r omega and tangential acceleration, while distinguishing tangential velocity from centripetal acceleration and angular quantities from their linear counterparts.
Practise Rotational and Circular Motion
521 free Rotational and Circular Motion MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Angular and Linear Quantities questions
If length of a spanner is 'I' and a force 'F' is applied on it to tighten a nut such that it passes through the pivot point, then torque is:
What is torque 'τ' in a circular motion?
Linear velocity or tangential velocity of any particle moving in a circular path of radlus 2m with angular velocity 8 rads-1 will be:
A force 'F' is acting at point 'P' of a uniform rod capable of rotating about 'O'. What is the torque about 'O'?
The ratio of moment of inertia of two objects 'A' and 'B' is 2:3. Which one of the following is the ratio of torques of 'A' and 'B' respectively if both are being rotated with constant angular acceleration?
If we double the moment arm, the value of torque becomes: