Question of Physics

Question

A block of mass m is released at rest at height h. It is undergoing

a circular motion around a loop of radius R represented by a dash line (without

falling off at the top of the loop). Find

a) the speed of the block at point A (at this position, the normal force on the

block is directed horizontally) and

b) the force acting on the block.

Answer 1

I am assuming that this question came for chegg.com and the quantities

you mention refer to Figure 2 there.

If I am wrong then ignore my answer because I do not understand the question.

a)

Let v be the velocity at point A.

The velocity v can be determined from conservation of energy.

The initial potential energy gets converted to kinetic energy at point A.

We can chose the point A to be the point of 0 potential energy;

the initial potential energy is then mg(h-R).

The kinetic energy at point A is mv²/2.

From their equality we can calculate v:

mv²/2 = mg(h-R) (1)

v = sqrt(2g(h-R))

Please note that v will be undefined if the quantity under the square root is negative,

i.e., h < R, which happens if the point A is higher than the initial starting point.

Actually the block will fail to behave as described unless it can reach the top of

the dashed circle with enough speed to overcome gravity.

However, the statement of the problem does not seem to ask for the domain of validity.

b)

There are two forces acting on the block

- downward force of gravity, mg

- horizontal centripetal force, mv²/R, which can rewritten using (1) as 2mg(h-R)/R.

The total force acting on the block is the vector sum of the two.

Since those two forces are perpendicular to each other, we can get the magnitude

of their sum using the Pythagorean theorem:

F = √((mg)² + (2mg(h-R)/R)²) = mg√(1 + (2(h-R)/R)²)