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/s2, where x= _____ (use g=10 m/s2).
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.
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