Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass MMM and radius RRR about an axis perpendicular to th
1 answer:
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When driver see the child standing on road his speed is 20 m/s
So here at that instant his reaction time is 0.80 s
He will cover a total distance given by product of speed and time
now after this he will apply brakes with acceleration a = 7 m/s^2
so the distance covered before it stop is given by
so the total distance covered by it
<em>so it will cover a total distance of 44.6 m</em>
Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x =
x =
let's calculate
x = 2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m