"An athlete whirls a 3.7kg shot-put in a horizontal circle with a radius of 0.90m. If the period of rotation is 0.30s, how far would the shot-put travel horizontally if it is released 1.2m above the level ground?"
There were two other parts to the question that I had already calculated. The speed of the shot-put when released is 18.85m/s, and the centripetal force acting on the shot-put is 1461N.
Thanks, but I got it.
To find the distance traveled by the shot-put horizontally, we can use the formula for the distance traveled in circular motion, which is given by:
Distance = Speed × Time
In this case, the distance traveled by the shot-put is the same as its circumference since it is revolving in a circle. So we can rewrite the formula as:
Distance = Circumference
The circumference of a circle is given by the formula:
Circumference = 2π × Radius
In this problem, the radius of the circle is given as 0.9m. Therefore, we can calculate the circumference as:
Circumference = 2π × 0.9m
Next, we need to find the time it takes for one complete rotation of the shot-put, which is given in the problem as the period of rotation. The period is the time it takes for the shot-put to complete one full revolution. In this case, the period is given as 0.30s.
Now we can substitute these values into the formula for distance:
Distance = Circumference = 2π × 0.9m
Calculating the circumference:
Circumference = 2π × 0.9m = 5.6549m
Finally, we can calculate the distance traveled by the shot-put by multiplying the calculated circumference by the number of revolutions in the given time period:
Distance = Circumference × (Number of Revolutions)
Since the period given is the time for one complete rotation, we can find the number of revolutions in the given time period as:
Number of Revolutions = Time / Period
Number of Revolutions = 1.2m / 0.30s = 4 revolutions
Substituting this value:
Distance = 5.6549m × 4 revolutions = 22.6196m
Therefore, the shot-put would travel horizontally a distance of approximately 22.6196 meters if it is released 1.2 meters above the level ground.
To find the horizontal distance traveled by the shot-put, we need to calculate the horizontal component of its velocity using the formula:
v = d/t
where v is the horizontal velocity, d is the horizontal distance, and t is the period of rotation.
Given:
- speed of the shot-put when released (v) = 18.85 m/s
- period of rotation (t) = 0.30 s
First, we need to find the horizontal velocity of the shot-put.
Step 1: Find the angular velocity (ω).
Using the formula:
ω = 2π / t
where ω is the angular velocity and t is the period of rotation.
Substituting the given value:
ω = 2π / 0.30 s
ω ≈ 20.94 rad/s (rounded to two decimal places)
Step 2: Find the horizontal velocity (v).
Using the formula:
v = r * ω
where v is the horizontal velocity, r is the radius, and ω is the angular velocity.
Substituting the given values:
v = 0.90 m * 20.94 rad/s
v ≈ 18.84 m/s (rounded to two decimal places)
Now, we have the horizontal component of the shot-put's velocity.
Step 3: Calculate the horizontal distance (d).
Using the formula:
d = v * t
Substituting the values:
d = 18.84 m/s * 0.30 s
d ≈ 5.65 m (rounded to two decimal places)
Therefore, the shot-put would travel approximately 5.65 meters horizontally if it is released 1.2 meters above the level ground.