1.A mass is moving in uniform circular motion under the effects of a centripetal force. If the force is doubled and the radius is cut in half, by what factor will the speed change?

V^2/R is doubled, since the force doubles..

If R is also halved, V stays the same.
That allows V^2/R to double.

Thanks

To determine how the speed will change when the force is doubled and the radius is halved, we need to consider the relationship between centripetal force, speed, and radius in uniform circular motion.

The centripetal force is given by the equation:

F = (mv^2) / r

where F is the centripetal force, m is the mass of the object, v is the speed, and r is the radius of the circular path.

Let's denote the initial values of the centripetal force, speed, and radius as F₁, v₁, and r₁. After the changes, the new values will be 2F₁, v₂, and r₂.

Since the force is doubled and the radius is halved, the new centripetal force becomes:

2F₁ = (m(v₂^2)) / (r₂/2)

From this equation, we can observe that the mass cancels out. Rearranging the equation, we have:

v₂^2 = 4F₁(r₂/2)

Next, we can express the initial speed in terms of the initial centripetal force and radius using the equation F₁ = (m(v₁^2)) / r₁:

v₁^2 = F₁(r₁/m)

The m's cancel out, and we can rewrite this as:

v₁^2 = F₁r₁

Now, let's divide the equation for v₂^2 by the equation for v₁^2:

(v₂^2) / (v₁^2) = (4F₁(r₂/2)) / F₁r₁

The F₁ term cancels out, resulting in:

(v₂^2) / (v₁^2) = 2(r₂ / r₁)

Taking the square root of both sides of the equation, we get:

v₂ / v₁ = √(2(r₂ / r₁))

Therefore, the speed will change by a factor of √(2(r₂ / r₁)).

Note: This factor depends on the ratio of the final radius to the initial radius, as indicated by (r₂ / r₁).