A 18 kg rock is attached to a 52 N/m spring, you then start to swing the spring above your head in a horizontal circle at 32 radians/s. Calculate how far the spring will stretch during this motion assuming its relaxed length is 1.5meters. Ignore any effects of gravity
force=Mass*w^2 * radius=18^32^2 *1.5
lengthnew=force/k=(18*(32^2*1.5)/52= ...
is that all there is to it?
To calculate the distance the spring will stretch during the motion, we can use the concept of centripetal force.
First, let's determine the magnitude of the centripetal force acting on the rock.
The centripetal force (Fc) is given by the formula:
Fc = m * ω^2 * r
where:
m = mass of the rock (18 kg)
ω = angular velocity (32 radians/s)
r = radius of the circular path (unknown)
Since the spring is attached to the rock and it is stretched, the tension in the spring provides the necessary centripetal force.
So, the magnitude of the tension force (T) in the spring is equal to the centripetal force:
T = Fc = m * ω^2 * r
Next, let's determine the displacement of the spring when it stretches.
Hooke's Law states that the force exerted by the spring is proportional to the displacement from its equilibrium position. In this case, the equilibrium position is the relaxed length of the spring.
The force exerted by the spring (Fs) is given by the formula:
Fs = k * Δx
where:
k = spring constant (52 N/m)
Δx = displacement from the relaxed length (unknown)
Since the tension force in the spring (T) provides the necessary centripetal force, we can equate the magnitudes of the tension force and the force exerted by the spring:
T = Fs
m * ω^2 * r = k * Δx
Rearranging the equation to solve for Δx:
Δx = (m * ω^2 * r) / k
Now, substitute the given values into the equation:
Δx = (18 kg * (32 radians/s)^2 * r) / 52 N/m
Simplifying the equation:
Δx = (18 kg * 1024 radians^2/s^2 * r) / 52 N/m
Δx = (18432 radians^2 kg/m * r) / 52 N
Finally, we can calculate the value of Δx using the given values.
Please note that the unit of the displacement (Δx) will be the same as the unit of the radius (r) in order to cancel out the unit of N (newtons) in the numerator and the denominator.
Given that the relaxed length of the spring is 1.5 meters, we can substitute this value into the equation:
Δx = (18432 radians^2 kg/m * 1.5 m) / 52 N
Simplifying further,
Δx = 55344 radians^2 kg / 52 N
Thus, the distance the spring will stretch during this motion is approximately 1063.4 meters.