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A cylinder is rolling down on an inclined plane of inclination 60∘. It's acceleration during rolling down will be x3√ m/s2, where x= _____ (use g=10 m/s2).

Question

A cylinder is rolling down on an inclined plane of inclination 60°.

It's acceleration during rolling down will be x3 m/s2x \sqrt{3}\ m/s^2, where x= _____ (use g=10 m/s2g=10\ m/s^2.

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Solution

The acceleration of a cylinder rolling down an inclined plane can be calculated using the formula:

a = g * sin(θ) / (1 + k^2 / r^2)

where:

  • g is the acceleration due to gravity,
  • θ is the angle of inclination,
  • k is the radius of gyration, and
  • r is the radius of the cylinder.

For a solid cylinder, the radius of gyration k is given by k = r / √2. Substituting this into the formula gives:

a = g * sin(θ) / (1 + 1/2)

Given that g = 10 m/s^2 and θ = 60 degrees, we can substitute these values into the formula to find the acceleration a:

a = 10 * sin(60) / (1 + 1/2) a = 10 * √3/2 / 1.5 a = 10√3 / 3 m/s^2

Therefore, the acceleration of the cylinder rolling down the inclined plane is 10√3 / 3 m/s^2. Comparing this with the given form x√3 m/s^2, we find that x = 10/3.

This problem has been solved

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