Solved Uniform Circular Motion Problems Physics

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Example 3: A block of mass \(2\) kilograms is sliding around in a circle at a constant speed of \(4\) meters per second.

If the coefficient of kinetic friction between the block and the ground is \(0.3\) and the constant acceleration of the block is \(10\) meters per second squared, find the radius of the circle formed by the block.

Consider a wet roadway banked, where there is a coefficient of static friction of 0.40 and a coefficient of kinematic friction of 0.2 between the tires and the roadway.

The radius of the curve is R=80m (a) if the banking angle is 30,what is the maximum speed the automobile can have before sliding up the banking?

You are driving along an empty straight road at a constant speed u.

Solved Uniform Circular Motion Problems Physics Example Of Business Plan Proposal

At some point you notice a tall wall at a distance D in front of you.What is the maximum number of revolutions per minute allowed if the string is not to break? The friction coefficient between the ruler and the block is .The ruler is fixed at one end and the block is at a distance L from fixed end in the horizontal plane through the fixed end.Solution: The fact that the speed, radius, and acceleration are constant mean that the block is undergoing uniform circular motion.There are two forces acting on the block on the plane of the ground: the centerpointing applied force on the block, and friction.In the film 2001: A Space Odyssey, a wheel like space station achieves artificial gravity by spinning around its axis.If the station had a size of 2 km, how fast should it be spinning for the people inside to feel the same gravitational acceleration as on earth?The stone has a mass of \(1\) kilogram, and a radius of \(12\) meters.The constant speed of the stone is \(6\) meters per second.The sum of the forces on this plane is $$\sum = F_R - F_$$ By Newton's Second Law: $$ma = F_R - F_$$ Now we know the formula for the applied force: $$ma = \frac - F_$$ We also know the formula for the frictional force: $$ma = \frac - mg\mu_k$$ Divide away \(m\): $$a = \frac - g\mu_k$$ Now solve for \(R\): $$a R = v^2 - Rg\mu_k \Rightarrow$$ $$a R Rg\mu_k = v^2 \Rightarrow$$ $$R(a g\mu_k) = v^2 \Rightarrow$$ $$R = \frac$$ Feel free to check the units on the right-hand side to ensure they are units of length.Now plug in the known values: $$R = \frac = 1.236 \; m$$ Example 4: A stone is rolling around in a uniform circular motion.

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