1. Describe what is happening at point A when these 3 vectors act on point A. 2. Describe what happen to the resultant vector at
Point B when the 5N vector goes from 0 degree to 180 degrees. 3. A lawnmower is pushed with a constant force F. as 0 between vector F and the horizontal plane decreases describe what is happening to the Fx and Fy vector components of vector F. 4.Calculate the magnitude of the component of the force parallel to the ground.
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