# Kinematic of rigid bodies

1. Jul 2, 2011

### CICCI_2011

What is solution to this problem any help is appreciated. Thanks

Around wheel that can rotate around a horizontal axis is coiled (reeled) light unstretchable rope whose other end is tied load P. The position of the body is determined at all time with coordinate z =(pi)t³/3. Determine current speed and angular acceleration of wheel at point in time after nine revolutions if the wheel radius is r = 4 cm. Determine the speed and acceleration of point B of lever end AB = L = 20 cm, which rotates together with the wheel.

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Last edited by a moderator: May 5, 2017
2. Jul 3, 2011

### ehild

Remember the rope does not stretch. How are the speed and acceleration of the load related to the speed and tangential acceleration of a point on the rim?

ehild

3. Jul 3, 2011

### CICCI_2011

There are the same but what are equations used in solving problem?

4. Jul 3, 2011

### ehild

Find the velocity and acceleration of the load first, from its displacement as function of time.

ehild

5. Jul 3, 2011

### CICCI_2011

Like this?

z =(pi)t³/3
z=(9*180)t³/3
z=540 t³
z'=Vp=3*540 t²

z''=ap=2*1620 t

angular speed ω=Vp/r=1620 t²/4

ω=405 t²

angular acceleration

ω'=3240 t/r

ω'=810 t

speed of B Vb=AB*ω

Vb=20*405 t²

acceleration of B
ab=AB*ω²

ab=20*(405t²)²

Is this true or if not how it's done. Help

6. Jul 3, 2011

### ehild

pi is the number pi=3.1415926....., you can not change it to anything. Determine both the v(t) and a(t) functions by derivation of z(t) with respect to time, using the original formula, z=(pi)t3/3.

During 9 revolutions, a length of rope L equal to 9 times the circumference of the wheel moved off. What is the circumference of the wheel? So the load attached to the rope moved down by this length, L, and z = L. Calculate the time needed to this from the formula L=(pi)t3/3 and plug in for t in the equations of velocity and acceleration.

ehild

7. Jul 3, 2011

### CICCI_2011

L=9*2rpi=9*2*4*3.14=226.08

226.08 =3.14t³/3

t=6 sec

z =(pi)t³/3

v(t)=(3(pi)t²)/3

v(t)=(3*3.14 6²)/3=113.04

a(t)=(3*2*(pi))t/3

a(t)=(2*2*3.14*6)/3=25.12

angular speed ω=v/r=113.04/4=28.26

angular velocity ω=a/r=25.12/4=6.28

Speed of B is Vb=ω*AB=28.26*20=565.2

Velocity of B is a=ω²*AB=6.28²*20=788.768

I this now OK? Thanks for your help

8. Jul 3, 2011

### ehild

The speed is the magnitude of the velocity. You need acceleration, do not mix it with velocity. The tangential acceleration divided by r gives the angular acceleration. The acceleration has both centripetal component and tangential one, which are perpendicular. The magnitude of the resultant acceleration is √(acp2+at2)
Take this into account when you determine the acceleration either of the rim, or of point B.

And do not forget to write out the units.

Otherwise your work is basically all right now.

ehild

Last edited: Jul 3, 2011
9. Jul 3, 2011

### CICCI_2011

I will mark first derivative with ' so ω'-angular acceleration and ω-angular speed.

So derivation of z(t) with respect to time gives speed of load and acceleration of load which is tangential acceleration of wheel. From tangential acceleration I can find angular acceleration of wheel ω'=a/r. And from speed of load I can find angular speed ω=V/r. After that I can find centripetal component as a=rω². Then I can find magnitude of the resultant acceleration as √(acp²+at²). For B speed is Vb=ωAB and acceleration is

tangential acceleration a=ABω'
centripetal acceleration a=ABω²

Is this true?

I'm not good with special characters so sorry for this kind of typing.

10. Jul 3, 2011

### ehild

It is all right, well done!

ehild

11. Jul 3, 2011

### CICCI_2011

Thanks a lot for your time and effort

Best regards