A 4kg mass moves in a circular path of 7m radius on a frictionless horizontal table. It is attached to a string that passes through a frictionless hole in the center of the table. A second 87kg mass is attached to the other end of the string. Acceleration is 9.8 m/s^2. Determine the period.
To determine the period of the system, we can use the formula for the period of a mass on a string in circular motion:
T = 2π√(r/g)
where T is the period, r is the radius of the circle, and g is the acceleration due to gravity.
In this case, the acceleration given (9.8 m/s^2) is the acceleration due to gravity (g), and the radius of the circular path is 7m.
Let's substitute these values into the formula:
T = 2π√(7/9.8)
Calculating this, we get:
T = 2π√(0.714)
Simplifying further, we have:
T = 2π(0.846)
Finally, calculating the value of T, we get:
T ≈ 5.34 seconds
Therefore, the period of the system is approximately 5.34 seconds.
To determine the period of the circular motion, we can start by analyzing the forces acting on the system.
First, we have the gravitational force acting on each object. The force on the 4kg mass is given by:
F₁ = m₁ * g = 4kg * 9.8 m/s^2 = 39.2 N
The force on the 87kg mass is given by:
F₂ = m₂ * g = 87kg * 9.8 m/s^2 = 852.6 N
Next, we have the tension force in the string. The tension force is responsible for providing the centripetal force required to keep the 4kg mass in circular motion. The centripetal force is given by:
F_c = m₁ * a = 4kg * 9.8 m/s^2 = 39.2 N
Since the tension in the string is the force that provides the centripetal force, we can equate the tension force to the centripetal force:
F_tension = F_c = m₁ * a
So, the tension force is 39.2 N.
Now, we can use the tension force and the acceleration to calculate the angular velocity (ω) of the 4kg mass:
F_tension = m₁ * a
39.2 N = 4kg * ω² * r
39.2 N = 4kg * (2π / T)² * 7m
39.2 N = (8π² * 4kg * 7m) / T²
T² = (8π² * 4kg * 7m) / 39.2 N
T² = 8π² * 4kg * 7m / 39.2 N
T² ≈ 22.34 s²
Finally, we can solve for the period (T) by taking the square root of both sides:
T ≈ √(22.34 s²)
T ≈ 4.72 s
Therefore, the period of the circular motion is approximately 4.72 seconds.
4 v^2/7 = 87 *9.8
solve for v
then
Time around circle = 2 pi*7/v