Given: $m,R,I=mR^2/2, F,M$ and no traction between $B_2$ and the ground find $a_1, a_2$ (the accelerations of the CMs of $B_1$ and $B_2$ respectively).$B_1$ rolls on $B_2$ without sliding due to $T$.
Asked
Active
Viewed 1,050 times
1
bolzano
- 196
-
How is the $a_1$ coming into your equation? – BowlOfRed Dec 10 '14 at 16:49
-
It is the acceleration of the center of mass of Body 1. – bolzano Dec 10 '14 at 16:58
-
Yes, but I mean it looks to me like you are bringing it in due to a relationship with the angular acceleration. Is that correct? What is that relationship? – BowlOfRed Dec 10 '14 at 17:00
-
$\alpha_{\gamma} = a_1 / R$ since it RWS – bolzano Dec 10 '14 at 19:04
-
That's only true in a reference frame where the axis of rotation is fixed. If the cylinder were stuck to body 2 and not rotating at all, $\alpha_{\gamma}$ would be zero, but $a_1$ positive. – BowlOfRed Dec 10 '14 at 19:27
-
And if you use a an accelerating reference frame (such as one where the axis of the cylinder is at rest), then you do get fictitious forces like your $F'$ – BowlOfRed Dec 10 '14 at 19:32
-
Hmmm, do you mean that it's a problem of relative motion? – bolzano Dec 10 '14 at 19:41
-
Do you realize that you can write your solution as an "answer"? – Floris Dec 13 '14 at 04:36
-
I've done it! Does this, however, assist to voting up? – bolzano Dec 17 '14 at 19:02
3 Answers
1
No, what would be the source of that force? there is no such force F', unless the problem explicitly put it as an external force. T is responsible for both the torque for rotation and the acceleration of the center of mass of the ball. Given the specific moment of inetria the ball has, you get a set of apparently contradictory equations. The solution is that is consistent will be T=0, which means that this specific ball cannot roll without sliding.
UPDATE: we can make the ball rolling if we apply a force F'. Ib this case there are two solutions to the problem. You can choose to put the force F' in the same direction of F, or in the opposite one. In the first case, the solution is $F'=.5ma_1=T$. In this example both T and F' are in the same direction. In the second solution both F' and T have the same direction, but this time opposite to F. The acceleration is negative. and in this case $F'=T=.5 ma_1$
in the first case, the solution for a1 is: $F+T=Ma_2$, thus: $a_2=(F+F')/M$
In the second case you have: $F-T=Ma_2$, thus: $a_2=(F-F')/M$
I am not sure how you got $a_2 = \frac{F}{m+3M}$
-
However if there is no $F'$ then $T=ma_1$ and $T=ma_1/2$ which is not true! – bolzano Dec 10 '14 at 16:03
-
I dont know how you got to the second equation. If it is supossed to the the torque it should be $TR=Ia/R$, thus you find a from the two conditions – Dec 10 '14 at 16:08
-
$$\sum \tau = I \alpha_{\gamma} \iff TR = 1/2mR^2 a_1/R \iff T = ma_1 / 2$$ – bolzano Dec 10 '14 at 16:09
-
Simulating, however, the experiment using Phun implies something totaly different (which is the existence of static traction) – bolzano Dec 10 '14 at 16:25
-
I might be doing something wrong here, let me solve the problem entirely. I'll be in touch – Dec 10 '14 at 16:29
-
where that statics fraction comes from? If you add that as an external force I agree, now there is a solution. But the original problem did not say there was an additional traction force (it cannot appear from the vaccum) – Dec 10 '14 at 16:33
-
Then why does the ball roll? There is a torque and therefore static traction – bolzano Dec 10 '14 at 16:38
-
yes, but the acceleration will not mach the right amount of rotation, so it will actualy slide. Unless toy add the external force in red that allow an acceleration ond angular acceleration consistent with the assumption of no-sliding – Dec 10 '14 at 16:51
-
-
So F' exists (this matches the data of the problem; that it must roll without sliding). Is the drawing however correct? – bolzano Dec 10 '14 at 16:57
-
The drawing could be correct. I need to solve the problem to see if F' is more of less that T, in the last case the acceleration of the ball should change direction (but I do not think intuitively that F' will be larger that T). Try to solve it yourself, I gotta go now but solve it tonight when I am back. – Dec 10 '14 at 17:01
-
$F' < T < F$ because it is going to accelerate at the positive (right) direction so $F' \neq F$ – bolzano Dec 10 '14 at 19:03
-
I was thinking of something like this and that's the reason I drew $F'$ with a red dashed arrow – bolzano Dec 10 '14 at 19:43
1
The center of mass of an object accelerates as though all forces acting on that object act there; there is no need to invoke the "red force" in your diagram. The torque on the cylinder is given by $\Gamma = T\cdot R$ and of course the difference between $T$ and $F$ is the force that accelerates the lower block, e.g. $F-T = M\cdot a_2$; make sure that the sign conventions are what you want them to be (you draw one $T$ to the right and another to the left...)
Floris
- 118,905
0
UPDATE (Solution):
The center of mass of $B_1$ has a displacement of $$\vec x_1 = \vec x_{1,1} + \vec x_{1,2}$$ One due to RWS and the other due to its contact with the other body. Differentiation (twice) leads to $$\vec a_1 = \vec a_{1,1} + \vec a_{1,2} \xrightarrow{algebraic} a_1 = a_{1,2} - a_{1,1}$$. But $$T = m a_1$$ and $$\sum \tau = I \alpha_{\gamma(1)} \iff T = ma_{1,1}/2$$ Finally $$F-T = ma_2 = m a_{1,2}$$ combining the equations leads to $$a_1 = \frac{F}{m+3M}$$ and $$a_2 = \frac{3F}{m+3M}$$
bolzano
- 196