Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
Actually, the speed of the earth is the same everywhere, taking the angular speed as the valid measure of the speed
Answer:
The vertical distance is ![d = \frac{2}{k} *[mg + f]](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B2%7D%7Bk%7D%20%2A%5Bmg%20%2B%20f%5D)
Explanation:
From the question we are told that
The mass of the cylinder is m
The kinetic frictional force is f
Generally from the work energy theorem
Here E the the energy of the spring which is increasing and this is mathematically represented as
Here k is the spring constant
P is the potential energy of the cylinder which is mathematically represented as
And
is the workdone by friction which is mathematically represented as
So
=> ![\frac{1}{2} * k * d^2 = d[mg + f ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%20%2A%20%20d%5E2%20%3D%20%20d%5Bmg%20%2B%20%20f%20%20%20%20%5D)
=> ![\frac{1}{2} * k * d = [mg + f ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20k%20%20%2A%20%20d%20%3D%20%20%5Bmg%20%2B%20%20f%20%20%20%20%5D)
=>
