A boy whirls a stone in a horizontal circle of radius 1.3 m and at height 2.0 m above level
ground. The string breaks, and the stone flies off horizontally and strikes the ground after
traveling a horizontal distance of 9.8 m. What is the magnitude of the centripetal acceleration
of the stone while in circular motion?
To find the magnitude of the centripetal acceleration of the stone while in circular motion, we can use the formula for centripetal acceleration:
ac = (v^2) / r
where ac is the centripetal acceleration, v is the velocity of the stone, and r is the radius of the circular motion.
In this case, we are given the horizontal distance traveled by the stone, which we can consider as the circumference of a circle with radius 1.3 m. Thus, we can use this information to find the velocity of the stone.
The distance traveled along the circumference of the circle is given by:
2πr = 9.8 m
Solving for the radius:
r = 9.8 m / 2π
r = 1.56 m
Now, we can use the formula for the velocity of an object in circular motion:
v = ωr
where v is the velocity, ω is the angular velocity, and r is the radius.
Since the stone breaks free and travels horizontally when it hits the ground, its velocity is purely horizontal. Therefore, the angular velocity is zero.
v = ωr = 0 * 1.3 = 0 m/s
Plugging this value into the formula for centripetal acceleration:
ac = (0^2) / 1.3
ac = 0 m/s^2
Therefore, the magnitude of the centripetal acceleration of the stone while in circular motion is 0 m/s^2.
To find the centripetal acceleration of the stone while in circular motion, we can use the formula:
Centripetal acceleration (a) = (v^2) / r
where:
- a is the centripetal acceleration,
- v is the velocity of the stone while in circular motion, and
- r is the radius of the circular motion.
In this case, we are given the radius of the circular motion (r = 1.3 m) and the distance traveled horizontally (d = 9.8 m). We need to find the velocity (v) to calculate the centripetal acceleration.
The distance traveled by the stone is the circumference of the circle, which can be calculated using the formula:
Circumference = 2πr
So, 2πr = d
Substituting the given values:
2π(1.3 m) = 9.8 m
Simplifying:
2π ≈ 9.8 / 1.3
2π ≈ 7.538
Now, we can calculate the velocity (v) of the stone using the formula:
v = d / t
where t is the time taken to travel the distance horizontally.
We know that the stone travels horizontally, so there is only horizontal velocity (no vertical velocity).
The time taken (t) can be calculated using the formula:
t = d / v
Substituting the given values:
t = 9.8 m / v
Now, we can substitute this value of t into the equation for v:
v = 9.8 m / (9.8 m / v)
Simplifying:
v = v
So, we can see that the velocity (v) of the stone remains unchanged.
Now, we can substitute the values of v and r into the formula for centripetal acceleration:
a = (v^2) / r
a = (v × v) / r
Substituting the given values:
a = (v × v) / 1.3 m
Since v remains unchanged, we can simplify this expression to:
a = (v^2) / 1.3 m
Therefore, the magnitude of the centripetal acceleration of the stone while in circular motion is (v^2) / 1.3 m.