# Angular velocity of two connected pulleys (halved velocity)

I have a problem with this exercise.

The weight A is going up with the velocity of 18 cm/s the fixed pulley has the radius of 6cm and the moving one of 3cm. What's the angular velocity of the moving one? Consider the rope does not slide on the pulleys

What I thought was the following: Since the rope is going up with the speed of 18cm/s then it's going down with the same speed. Using the formula $$W = \frac{(v1-v0)}{r}$$ being v1 and v0 two velocities in the pulley and r the distance between them. I did the following $$W = \frac{18-0}{3}$$ which equals 6. But the answer is 3.

I know why the answer is 3 but can't understand. $$Length = ya + 2*yb$$ Derivating we get $$vb = -va/2$$

Applying vb on the same formula I indicated before we get $$W = 9/3$$ which is in fact equal to 3. But Vb is not the velocity of the pulley but the block B.

My question here is why does the velocity of the rope running on the moving pulley is half of the velocity of the rope running on the fixed one, it should be the same. According to the exercise both pulleys have the same angular velocity.

If you take this example. The smaller one has a bigger angular velocity.

But according to this exercise they can have the same velocity.

• The standard unit for angular velocity is $\omega$. Please use '\omega' to write $\omega$ instead of $W$ for angular velocity. – Dlamini Jun 11 '18 at 0:00

I know why the answer is 3 but can't understand.

I am afraid that I do not.

If the rope is inextensible then the linear speed of every part of the rope must be the same.

If there is no friction between the pulley wheels and the rope then the pulley wheels will not rotate.

If there is sufficient friction between the pulley wheels and the rope such that there is not relative movement between the edge of a pulley wheel and the rope then the linear speed of the outer rim of a pulley wheel must equal the speed of the rope.
$$\text{angular speed of a pulley}= \dfrac {\text{linear speed of the rope}}{\text{radius of the pulley wheel} }$$

The linear speed of block $B$ is related to the linear speed of block $A$ and the arrangement of pulley wheels.
In this case if you raise block $A$ by $16\, \rm m$ the block $B$ will fall by $8 \, \rm m$.

• Sorry this problem was not in english so I tried to translate it. What I mean is that the rope doesn't slide on the pulley. – krystalgamer Jun 10 '18 at 12:24
• The part of the block B falling at the speed of 8cm/s I understood but I can't understand how does the speed of block B relates to the pulley. – krystalgamer Jun 10 '18 at 12:26
• @krystalgamer The downward speed of block $B$ is the same as the downward speed of the centre of mass of the bottom pulley. The linear speed of the edge of that pulley is the same as the speed of the rope. – Farcher Jun 10 '18 at 12:43
• @krystalgamer I do not think that $3$ is the correct answer for the angular speed. You have a translation of the centre of mass of the pulley wheel plus a rotation of the pulley wheel. The centre of mass of pulley wheel $B$ moves down at $\frac{18}{2} =9\, \rm cm\, s^{-1}$ and its angular speed is $\frac{18}{3}= 6\, \rm rad\, s^{-1}$. – Farcher Jun 10 '18 at 13:08

The rope does not slide on the pulley. So:

         υΕ = 0  →  υΒ = υΡ            (1)


Also: υA = υZ = υΒ + υΡ (2)

(1),(2): υA = 2υΒ → υΒ = 0,09 m/s and ω = 0,09/0,03 = 3 r/s

The way it has been solved is according to the concept of absolute motion dependent on weight. Let me answer your confusion first. You said the angular velocity must be same, it's true, you are right but with a condition. Here, angular velocity seems not the same because the rope should connect both pulleys alone. I mean, one of the sides of pulleys is connected with rope while the other side is directly connected to weight and fixed point respectively. So, there is not possible to have angular velocity the same. You solve it by considering the center points of pulleys as the datum. Mathematically, ya + yb + (yb-6cm) = total lenght of rope(it is constant term)

if you differentiate it with respect to time, you get, or, Va + 2*Vb = 0 or, Vb = -Va/2
so, Vb = -9 cm/s (-) sign indicates Vb going downward direction

angular velocity = Vb/radius = 9/3 = 3