This set of questions uses an interesting formula. A torque applied for a certain time causes a change in angular momentum.
Torque * (delta t) = delta L
Lindsey is on the merry-go-round again. Her mass is 33kg . The merry-go-round has a mass of 78kg and a radius of 2.20m. Lindsey is standing 0.150m from the center and has an initial angular velocity of 3.45 rad/sec. Her older brother Mike applies a force of 200 N tangent to the outer edge for causing the merry-go-round to spin faster. What was the initial angular momentum before Mike pushed?
I'm lost in the information...what steps do I need to take, and how does that formula factor in?
torque = dL/dt = rate of change of angular momentum
What else is new?
I is moment of inertia of merry-go-round with Lindsey on it.
omega = 3.45
L = I omega = 3.45 I
that is the initial angular momentum
Torque = I d omega/dt
or
200 * 2.20 = I d omega/dt
if you know I you can get d/dt (omega)
Thank you so much!!
To find the initial angular momentum before Mike pushed, we can use the formula:
angular momentum (L) = moment of inertia (I) * angular velocity (ω)
The moment of inertia depends on the distribution of mass around the axis of rotation. For a solid disk like the merry-go-round, the moment of inertia is given by:
moment of inertia (I) = (1/2) * mass * radius^2
Let's calculate the moment of inertia for the merry-go-round:
I = (1/2) * 78 kg * (2.20 m)^2
I = 178.86 kg·m^2
Now, we have the moment of inertia (I) and the initial angular velocity (ω) given as 3.45 rad/sec. We can calculate the initial angular momentum (L):
L = I * ω
L = 178.86 kg·m^2 * 3.45 rad/sec
L ≈ 617.88 kg·m^2/s
Therefore, the initial angular momentum before Mike pushed is approximately 617.88 kg·m^2/s.