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
the spring provides the centripetal acceleration ... keeps the rock moving in a circle
f = k x = mv^2 / r ... v = 32 r
r = x + 1.5
k x r = m (32 r)^2 ... k x = 32^2 m (x + 1.5)
solve for x
why is it in the third line, when you moved 'r' over to the other side, you multipled? at
'k x = 32^2 m (x + 1.5)'. wouldn't it be
k x = 32^2 m / (x + 1.5)?
there was an r^2 on the right
... one of them cancelled with the r from the left
thank you this helped alot
To calculate how far the spring will stretch during this motion, we need to use Hooke's Law and centripetal force.
1. First, let's calculate the force exerted by the spring when stretched. According to Hooke's Law, the force exerted by a spring is given by F = k * x, where F is the force, k is the spring constant, and x is the displacement from the relaxed position. In this case, the spring constant is 52 N/m and the displacement is unknown.
2. We know that the force exerted by the spring is equal to the centripetal force, which is given by F = m * w^2 * r, where m is the mass, w is the angular velocity, and r is the radius of the circular motion.
3. Rearranging the equations, we have k * x = m * w^2 * r.
4. We can substitute the given values: k * x = (18 kg) * (32 rad/s)^2 * r.
5. We also know that the relaxed length of the spring is 1.5 meters, so the total length of the spring when stretched is 1.5 meters plus the displacement x.
6. Solving for x, we get x = [(18 kg) * (32 rad/s)^2 * r] / (52 N/m).
7. The total length of the spring when stretched is 1.5 meters + x.
By plugging in the values and solving for x, we can find the distance the spring will stretch during this motion.