A child sits down on one end of a horizontal seesaw of negligible weight, 2.14 m from the pivot point. No one balances on the other side. Find the instantaneous angular acceleration of the see-saw and the tangential linear acceleration of the child.
the tangential linear acceleration of the child. = g = 9.81m/s^2
instantaneous angular acceleration of the see-saw = a/r = 9.81/1.91 = 5.14rad/s^2
I know that the angular acc is g/d or 9.8/2.14=4.58 rad/s^2
where is the 1.91 coming from?
opps, switch that with the 2.14m
To find the instantaneous angular acceleration of the seesaw, we need to use the equation:
α = τ / I,
where α is the angular acceleration, τ is the torque, and I is the moment of inertia.
In this case, since there is no one balancing on the other side, the seesaw is not in equilibrium. The child's weight creates a torque that will cause the seesaw to rotate.
The torque can be calculated as:
τ = r * F,
where r is the distance from the pivot point to the point where the force is applied (in this case, the child's position), and F is the force acting at that point.
In this case, the child's weight is the force acting on the seesaw, and it can be calculated as:
F = m * g,
where m is the mass of the child, and g is the acceleration due to gravity. We can assume that the mass of the child is given.
Now we can calculate the torque:
τ = 2.14 m * (m * g).
Next, we need to determine the moment of inertia of the seesaw, which depends on its shape and distribution of mass. In this case, since the seesaw is described as having negligible weight, we can assume that its moment of inertia is also negligible.
Therefore, the moment of inertia, I, can be considered as zero.
Now we can calculate the angular acceleration:
α = τ / I = (2.14 m * m * g) / 0 = undefined.
The angular acceleration is undefined in this case because there is no moment of inertia for the seesaw. This indicates that the seesaw will not rotate in response to the child's weight.
As for the tangential linear acceleration of the child, since the child is sitting at one end of the seesaw, there is no motion along the tangent direction. Therefore, the tangential linear acceleration of the child is zero.