All Exams Test series for 1 year @ ₹349 only
Question

A ball rotates at a rate r rotations per second and simultaneously revolves around a stationary point O at a rate R revolutions per second (R < r). The rotation and revolution are in the same sense. A certain point on the ball is in the line of the centre of the ball and point O at a certain time. This configuration repeats after a time

The correct answer is \(\frac{1}{r-R}\)

Ball Rotation and Revolution

The problem describes a ball that is simultaneously rotating about its own center and revolving around a stationary point. Both motions occur in the same direction (sense). We are asked to find the time after which a specific point on the ball, which is initially in the line connecting the stationary point O and the center of the ball C, repeats this configuration.

Angular Speeds

Let the rate of rotation of the ball be $r$ rotations per second. The angular speed of rotation, $\omega_r$, is related to the rotation rate by: $$ \omega_r = 2\pi r \text{ radians per second} $$ Let the rate of revolution of the ball's center around the stationary point O be $R$ revolutions per second. The angular speed of revolution, $\omega_R$, is related to the revolution rate by: $$ \omega_R = 2\pi R \text{ radians per second} $$ Both the rotation and revolution are in the same direction.

Relative Angular Speed

Consider a specific point P on the surface of the ball. This point P rotates about the center of the ball C. From the perspective of a fixed point in space, the line segment CP rotates with an angular speed $\omega_r$. The center of the ball C revolves around the stationary point O. The line segment OC (connecting O and C) revolves around O with an angular speed $\omega_R$. Initially, the point P is in the line of O and C. Let's assume P is on the side of C away from O (O-C-P). For this configuration to repeat, the point P must again be in the line OC in the same relative position. This means the line segment CP must align with the line segment OC in the same relative orientation.

We can analyze the motion of the line segment CP relative to the line segment OC. Since both are rotating/revolving in the same direction, and the rotation rate $r$ is greater than the revolution rate $R$ (given as $R < r$), the line segment CP is rotating faster than the line segment OC is revolving. The relative angular speed of the line segment CP with respect to the line segment OC is the difference between their absolute angular speeds: $$ \omega_{rel} = \omega_r - \omega_R $$

Time for Configuration to Repeat

For the configuration (point P being on the line OC in the same relative position) to repeat, the line segment CP must complete a full rotation (or an integer multiple of full rotations) relative to the line segment OC. The smallest time for the configuration to repeat is when the relative angle covered is $2\pi$ radians.

Let $T$ be the time taken for the configuration to repeat. The relative angle covered in time $T$ is $\omega_{rel} T$. For repetition, this relative angle must be $2\pi$: $$ \omega_{rel} T = 2\pi $$ Substitute $\omega_{rel} = \omega_r - \omega_R$: $$ (\omega_r - \omega_R) T = 2\pi $$ Now substitute the expressions for $\omega_r$ and $\omega_R$ in terms of rates $r$ and $R$: $$ (2\pi r - 2\pi R) T = 2\pi $$ Factor out $2\pi$: $$ 2\pi (r - R) T = 2\pi $$ Divide both sides by $2\pi$: $$ (r - R) T = 1 $$ Solve for $T$: $$ T = \frac{1}{r-R} $$

This is the time required for the specific point on the ball to be back in the line of the center of the ball and point O, repeating the initial configuration.

Was this answer helpful?

Important Questions from Memory Management

  1. A system has a Translation Lookaside Buffer (TLB) that has a reach of 1 MB. TLB reach is defined as the total amount of physical memory that can be accessed through the TLB entries. The paging system uses pages of size 4 KB. The virtual address space is 64 GB and physical address space is 1 GB. If each TLB entry stores a 4-bit process id, page number, frame number, and a 2-bit control field, then the size of the TLB (in bytes) is ___________. (answer in integer)

    Note: 1K=$2^{10}$, 1M=$2^{20}$, 1G=$2^{30}$
  2. Consider contiguous allocation of physical memory to processes using variable partitioning scheme. Suppose there are 8 holes in the memory of sizes 20 KB, 4 KB, 25 KB, 18 KB, 7 KB, 9 KB, 15 KB, and 12 KB. Assume that no two holes are adjacent. Two processes P1 of size 16 KB and P2 of size 9 KB arrive in that order, and they are allocated memory using the best-fit technique. After allocating space to P1 and P2, the number of holes of size less than 8 KB is ____________. (answer in integer)

    Note: 1K=$2^{10}$
  3. Match List I with List II
    List IList II
    A. Clustered Page TableI. Generally considered inappropriate for 64 bit architectures.
    B. Hierarchical Page TableII. Has only entry for each real page (or frame) of memory.
    C. SegmentationIII. Useful for Sparse address spaces.
    D. Inverted Page TableIV. Supports a user view of the system.

    Choose the correct answer from the options given below:
  4. Choose the correct answer.
    In memory management, page size is always, power of 2.
  5. Which of the following is a type of memory and is similar to RAM?

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App