A tether ball tied to a pole by a rope swings in a circular path with centripetal acceleration of 2.7m/s. If the ball has a tangential speed of 2.0m/s, what is the diameter of the circular path in which it travels?
1.5
To determine the diameter of the circular path in which the tether ball travels, we can use the formula for centripetal acceleration:
a = (v^2) / r
Where:
a = centripetal acceleration
v = tangential speed
r = radius or half the diameter of the circular path
Given:
a = 2.7 m/s^2 (centripetal acceleration)
v = 2.0 m/s (tangential speed)
First, we need to calculate the radius, so let's rearrange the formula:
r = (v^2) / a
Substituting the given values into the formula:
r = (2.0^2) / 2.7
r = 4.0 / 2.7
r ≈ 1.48 m
Finally, to determine the diameter, we double the radius:
Diameter = 2 * r
Diameter = 2 * 1.48
Diameter ≈ 2.96 m
Therefore, the diameter of the circular path in which the tether ball travels is approximately 2.96 meters.
To find the diameter of the circular path, we can use the formula relating centripetal acceleration, tangential speed, and radius:
ac = v^2 / r
where ac is the centripetal acceleration, v is the tangential speed, and r is the radius.
In this case, we are given that the centripetal acceleration is 2.7 m/s^2 (positive value since it has to be towards the center of the circle) and the tangential speed is 2.0 m/s.
Rearranging the formula, we get:
r = v^2 / ac
Substituting the given values, we have:
r = (2.0 m/s)^2 / 2.7 m/s^2
Solving this equation, we find:
r ≈ 1.48 m
The radius of the circular path is approximately 1.48 meters. Since the diameter is twice the radius, we can calculate the diameter:
d = 2r
Substituting the value of the radius, we have:
d = 2(1.48 m)
d ≈ 2.96 m
Therefore, the diameter of the circular path in which the tether ball travels is approximately 2.96 meters.
draw a force diagram. You have at the ball, mg downward, and outward hoizonal, m*2.7
but 2.7= v^2/r, you have v, so solve for r, thence Diameter.