A child is twirling a 0.0165-kg ball on a string in a horizontal circle whose radius is 0.149 m. The ball travels once around the circle in 0.543 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase?
v=1/period.
centforce= m*v^2/r
I am not certain why you are posting all these easy "plug and chug" problems tonight.
To determine the centripetal force acting on the ball, we can use the formula for centripetal force:
F = (m * v^2) / r,
where F is the centripetal force, m is the mass of the ball, v is the velocity of the ball, and r is the radius of the circle.
(a) To find the centripetal force:
First, we need to calculate the velocity of the ball. Since we know the radius and the time it takes for the ball to travel once around the circle, we can find the velocity using the formula:
v = (2 * π * r) / T,
where v is the velocity, π is a mathematical constant (approximately 3.14159), r is the radius, and T is the time period.
Substituting the given values:
v = (2 * 3.14159 * 0.149 m) / 0.543 s.
Simplifying:
v = 0.872 m/s.
Now, we can plug the values of mass, velocity, and radius into the centripetal force formula:
F = (0.0165 kg * (0.872 m/s)^2) / 0.149 m.
Calculating:
F = 0.0836 N.
Therefore, the centripetal force acting on the ball is approximately 0.0836 N.
(b) If the speed of the ball is doubled, we can find the factor by which the centripetal force increases. Let's call the new velocity v2.
We know that the centripetal force is proportional to the square of the velocity, so:
F2 / F1 = (v2 / v1)^2,
where F2 is the new centripetal force, F1 is the original centripetal force, v2 is the new velocity, and v1 is the original velocity.
If the speed is doubled, the new velocity v2 is 2 times the original velocity:
v2 = 2 * v1.
Substituting this into the equation:
F2 / F1 = ((2 * v1) / v1)^2.
Simplifying:
F2 / F1 = 4.
Therefore, if the speed is doubled, the centripetal force increases by a factor of 4.