Answer:
Therefore, the moment of inertia is:
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
(1)
Where:
- I is the moment of inertia
- M is the mass of the pendulum
- d is the distance from the center of mass to the pivot
- g is the gravity
Let's solve the equation (1) for I
![T=2\pi \sqrt{\frac{I}{Mgd}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7BI%7D%7BMgd%7D%7D)
![I=Mgd(\frac{T}{2\pi})^{2}](https://tex.z-dn.net/?f=I%3DMgd%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%29%5E%7B2%7D)
Before find I, we need to remember that
Now, the moment of inertia will be:
Therefore, the moment of inertia is:
I hope it helps you!
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 +
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
Answer:
i think it would be a but i am not 100% sure on that