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The following physics responses are provided by a student and, as such, should be considered in that context. While every effort is made for accuracy, absolute correctness cannot be fully guaranteed.
| Question | |
C | 8 |
I | |
Question A solid disk of radius R rolls down an incline in time T. The center of the disk is removed up to a radius of R/2. The remaining portion of the disk with its center gone is again rolled down the same incline. The time it takes is: | |
Answer more than T. | |
Notes
| User | Added | Note | e | d |
|---|---|---|---|---|
| KillGorack | 2023-06-29 17:51:48 | The moment of inertia of a solid disk is given by I = 0.5MR^2 where M is the mass and R is the radius. When the center of the disk is removed, we are left with a ring (annular disk) which has a moment of inertia given by I = MR^2. If the incline and the initial and final points of the motion are the same in both cases, then the difference in gravitational potential energy between these two points is the same and it's converted into kinetic energy. Kinetic energy of a rolling object consists of two parts: translational kinetic energy, which depends on the velocity of the center of mass (v_cm), and rotational kinetic energy, which depends on the angular velocity (ω). These are related by v_cm = ωR for rolling without slipping. In the case of the solid disk, the rotational kinetic energy is smaller than in the case of the annular disk (because 0.5MR^2 < MR^2 for the same mass and radius). This means that more of the potential energy gets converted into translational kinetic energy (and therefore speed) in the case of the solid disk. In other words, when the center is removed, less energy is available for linear motion down the ramp because more of it has to go into causing the ring to rotate. Therefore, the object with the center removed will roll slower than the full disk and it will take more time to reach the bottom of the incline. So the answer is 2) more than T. |
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