A horizontal force of 140 N is needed to pull a 60kg box across the horizontal floor at constant speed . Find the coefficient of
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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:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;
(ii) The linear speed of the small metal object is calculated as;