A child ties a rock to a string and whirls it around in a horizontal circle.
Assuming a 1.50 m -long string making 20.0 ∘ angle below the horizontal, find the speed of the rock.
Find the period of its uniform circular motion.
ok, use the 20 degrees.
tan20=vertical force/horizontal force
horizontal force=vertical force/tan20
mv^2/r=mg*ctn20
solve for v.
Tension = T
T sin 20 = weight = m g
so
T = m g/sin 20 = 28.7 * m if g = 9.81
m v^2/R = T cos 20 = 28.7 m (.94)
so
v^2/R = 27
v^2 = 27 R
R = 1.5 cos 20 = 1.41 meter
so
v^2 = 27(1.41)
v = 6.17 m/s
circumference = 2 pi R = 8.86 meters
time around = 8.86 meters/6.17 m/s
= 1.44 seconds
The answers given was v=6.36m/s and t=1.39s.
Any insight where I calculated wrong. I got basically the same as Damon but not rounded until the final steps.
To find the speed of the rock, we'll use the concept of centripetal force. Centripetal force is given by the equation:
F = (m * v^2) / r
Where F is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular motion.
In this case, the force is provided by the tension in the string. The gravitational force acting on the rock can be decomposed into two components: one along the string, balancing the tension force, and the other perpendicular to the string, providing the centripetal force.
The vertical component of the weight is given by:
F_vert = m * g * cos(theta)
Where g is the acceleration due to gravity (9.8 m/s^2) and theta is the angle below the horizontal (20.0 degrees).
Since F_vert balances the tension in the string, we can equate them:
F_vert = T = (m * v^2) / r
Substituting the value of F_vert and simplifying, we get:
m * g * cos(theta) = (m * v^2) / r
Simplifying further, we can solve for v:
v^2 = r * g * cos(theta)
Taking the square root of both sides, we find:
v = sqrt(r * g * cos(theta))
Now we can substitute the given values to find the speed of the rock.
Using r = 1.5 m, g = 9.8 m/s^2, and cos(theta) = cos(20 degrees) ≈ 0.93969, we can calculate:
v = sqrt(1.5 * 9.8 * 0.93969)
v ≈ 4.86 m/s
Therefore, the speed of the rock is approximately 4.86 m/s.
To find the period of its uniform circular motion, we can use the equation:
T = (2 * π * r) / v
Where T is the period, r is the radius of the circle, and v is the velocity.
Substituting the given values, we get:
T = (2 * π * 1.5) / 4.86
T ≈ 3.10 s
Therefore, the period of its uniform circular motion is approximately 3.10 seconds.