What is the momentum of a 110kg fullback running at 860 m/s? *
1 answer:
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Answer:
At t = 4.2 s
Angular velocity: 6. 17 rad /s
The number of revolutions: 2.06
Explanation:
First, we consider all the forces acting on the pulley.
There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.
Therefore, the torque on the pulley is
where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.
Now we also know that the torque is related to angular acceleration α by
therefore, equating this to the above equation gives
solving for alpha gives
Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives
Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.
The first kinematic equation we use is
since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have
Therefore, ar t = 4.2 s, the above gives
So how many revolutions is this?
To find out we just divide by 2 pi:
Or about 2 revolutions.
Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:
Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:
Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s
To summerise:
at t = 4.2 s
Angular velocity: 6. 17 rad /s
The number of revolutions: 2.06
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s
Power is defined as the rate at which the body is doing work:
Work is defined as displacement done by the force times that displacement:
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
of work.
Now we just divide that by 5s to get how much power is required:
Work is defined as displacement done by the force times that displacement:
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
of work.
Now we just divide that by 5s to get how much power is required: