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IgorLugansk [536]
2 years ago
5

Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri

ction) in the shape of a uniform disk with radius R = 7.50 cm and mass Mp = 14.1 kg. The string causes the pulley to rotate without slipping. If the masses are started from rest and allowed to move 2.50 m: What is the final speed (m/s) of mass m? What is the final angular speed (rad/s) of the pulley? How long (s) did it take for the masses to move from rest to the final position? Part (a) can be done by two methods: Forces and torques and energy conservation.
Physics
1 answer:
Scilla [17]2 years ago
5 0

Answer:

(a) 7 m/s

(b) 931 rad/s

(c) 0.716 s

Explanation:

Gravity would be exerting on the 2 masses

G_1 = Mg = 15.3*9.81 = 150.093N

G_2 = mg = 8.29*9.81= 81.32N

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

F = G_1 - G_2 = 68.77N

This force would generate torque on the solid pulley

T = FR = 68.77 * 0.0075 = 0.5158 Nm

We can also calculate the pulley moments of inertia, with it being solid

I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad

(c)The time it takes to get to this distance is

\theta = \frac{\alpha t^2}{2}

t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513

t = \sqrt{0.513} = 0.716s

(b)The final angular speed of the disk is

\omega = \alpha t = 1300.58*0.716 = 931 rad/s

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

v = \omega R = 931*0.0075 \approx 7m/s

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