A baseball pitcher throws a baseball horizontally at a linear speed of 41.3 m/s. Before being caught, the baseball travels a horizontal distance of 15.4 m and rotates through an angle of 52.7 rad. The baseball has a radius of 3.74 cm and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?
To find the tangential speed of a point on the "equator" of the baseball, we can use the rotational motion principles.
First, let's find the angular speed of the baseball. We know that the baseball rotates through an angle of 52.7 radians, and the time it takes to travel this distance is given by the equation:
θ = ω * t
Where:
θ is the angle in radians
ω is the angular speed in radians per second
t is the time in seconds
Rearranging the equation to solve for ω:
ω = θ / t
In this case, the angle is 52.7 radians, and we need to find the time it takes to travel the 15.4 m horizontally.
Since the baseball is thrown horizontally, the vertical component of its velocity is zero. Therefore, we can use the equation of motion:
d = vt
Where:
d is the horizontal distance traveled (15.4 m)
v is the horizontal velocity (41.3 m/s)
t is the time taken
Rearranging the equation to solve for t:
t = d / v
Substituting the values, we have:
t = 15.4 m / 41.3 m/s
Now we can substitute the values of θ and t into the equation to find ω:
ω = 52.7 rad / (15.4 m / 41.3 m/s)
Simplifying the equation, we get:
ω = (52.7 rad * 41.3 m/s) / 15.4 m
ω = 142.581 rad/s
Now we have the angular speed, we can find the tangential speed of the equator point. The tangential speed is given by:
v_tangential = ω * r
Where:
v_tangential is the tangential speed
r is the radius of the baseball
Substituting the values, we have:
v_tangential = 142.581 rad/s * 0.0374 m
Calculating, we find:
v_tangential = 5.33582 m/s
Therefore, the tangential speed of a point on the "equator" of the baseball is approximately 5.34 m/s.