Acceleration in terms of distance, but not time or final velocity?

[madchemist]

"The Hall car of mass m-c is connected to mass m-2 by a string as shown in the figure below. The string passes over a solid cylindrical pulley, which has a frictionless bearing of radius R and mass M. When the system is released from rest the string does not slip, the Hall car moves down the incline, and m-2 moves upwards through a distance h'. Derive an expression for the translational acceleration of the masses in terms of m-c, m-2, M, h', 2, and the coefficient of friction U." (I suspect that the term "2" might be a typo for theta; however, even with that assumption I can't solve in terms of h').

The figure is simply a right triangle with the car descending a distance h' along it's hypotenuse and m-2 ascending along it's vertical leg a distance h'. Theta is the angle opposite to the vertical leg and at the top of the vertical leg is the pulley. Finally, there is a friction arrow from the car directed opposite the direction in which the car is descending.


Tried these equations after finding the Net Force on the car, block and pulley: Torque= Tension times radius= Moment of Inertia times Angular Acceleration= (.5MR^2)(a/R)= (.5MRa).
Also PE= KE.



I was unsuccessful using the energy conservation approach, i.e., the PE of the car= the KE of the car and the rotational KE of the pulley. With the Newtonian approach: a= 2[(m-c)(g)(sin8)-(U)(m-c)(g)(cos8)-(m-c)(g)]/[M+(m-c)+(m-2)]. This solution is not in terms of h'. Moreover, it is in terms of the seemingly unavoidable g and theta.

 

[madchemist]

Judging by your silence, should I assume that there is no answer to this problem?

 

[Shooting Star]

For convenience of writing, let m-c=m1, m-2=m2, theta = b, u = co-eff of friction, w = omega of pulley. Let the tension in the string be T.

For m1 on the incline, m1*a = m1gsin b –T – um1gcos b ...(1)

For m2 ascending, m2*a = T –m2g …(2)

For pulley, torque = T*r = Idw/dt =I/r dv/dt = I*a/r, since v=rw => dw/dt = 1/r dv/dt = a/r, where v is the instantaneous velo of string and masses, and a= dv/dt. This gives
T = Ia/r^2 …(3) [where I = Mr^2/2]

Put value of T from 3 in 1 and 2 to get two eqns. Then eliminate g from them. You are left with the angle b still. I’m sure you can derive a geometric relation between h’ and angle b.

 

[madchemist]

Thanks for your response, however, I remain unable to eliminate g from the solution. I assume that since equatiions 1 & 2 are in terms of T (tension) that you mean to substitute T (tension) derived from equation 3 by dividing Mr^2/2*a/r by r and getting the tension equal to Ma/2. Taking the difference in Tensions from equations 1 & 2 and setting that equal to Ma/2 gives me a solution (w/o the mess of the geometric relation between h' and the angle b) sill in terms of g: 2[m1gsinb-um1gcosb-m2g]/(M+m1+m2).

 

[Doc Al]

Why are you trying to eliminate g? It's a constant. Your answer looks almost right.

 

[madchemist]

No need to eliminate g? What a relief! Anyway, here's what I have: a=2m1gsinb-2um1gcosb-2m1a-2m2a-2m2g=2(m1gsinb-um1gcosb-m2g)/(M+m1+m2). What am I missing?

 

[Doc Al]

How did you get your answer? (Show the three equations that you used to derive it.)

 

[madchemist]

You're right as always! How do you do this stuff so fast? Here's the corrrection: a=2m1gsinb-2um1gcosb-2m1a-2m2a-2m2g=2(m1gsinb-um1gcosb-m2g)/(M/2+m1+m2).

 

[Doc Al]

It's getting closer. (Please write it more carefully.)

 

[madchemist]

a=2m1gsinb-2um1gcosb-2m1a-2m2a-2m2g=(m1gsinb-um1gcosb-m2g)/(M/2+m1+m2). Note to self: Don't cut and paste! Is it right now?

 

[Doc Al]

OK, now you get a Scooby snack! (Yes, it looks correct.)

Careful how you write your work. When I see two things connected by an equals sign, I expect them to be equal--or at least have the same units. (Note that the terms in your middle expression do not have units of acceleration, so your first "equals" makes no sense.)

 

[Shooting Star]

Some more thoughts...

Reading the original problem once more, I noticed a number of discrepancies. It's given that the pulley is on frictionless bearing, but it's also given that there is friction between string and pulley, because the string does not slip. In that case the tensions in the two parts of the string are different. I had assumed both to be T while writing down the three equations.

Perhaps another attempt at the solution is called for...in terms of the quantities originally given by the OP.

 

[Doc Al]

Yes, the tensions are different. Although work wasn't shown, the answer given in post #10 looks correct to me.

 

[Shooting Star]

If he has done what I'd advised him to do in post no. 3, then the accn 'a' would also cancel out. And I didn't advise him to take different tensions. I'll try to do the whole calculation after some time.

Hey madchemist, please post the problem once more, without the typos. Remove the solved tag.

 

[madchemist]

Per your request Shooting star, and I quote: "The Hall car of mass mc is connected to mass m2 by a string as shown in the figure below. The string passes over a solid cylindrical pulley, which has a frictionless bearing of radius R and mass M. When the system is released from rest the string does not slip, the Hall car moves down the incline, and m2 moves upwards through a distance h'. Derive an expression for the translational acceleration of the masses in terms of mc, m2, M, h', 2, and the coefficient of friction U."

The figure is simply a right triangle with the car descending a distance h' along it's hypotenuse and m2 ascending along it's vertical leg a distance h'. Theta is the angle opposite to the vertical leg and at the top of the vertical leg is the pulley. Finally, there is a friction arrow from the car directed opposite the direction in which the car is descending. Note also that h is the vertical leg projection of h' with regard to the car.

 

[Doc Al]

...the Hall car moves down the incline, and m2 moves upwards through a distance h'.

I fail to see how the distance h' has anything to do with the acceleration.

 

[Shooting Star]

Thank you for reposting the problem. But what's the meaning of the "2"? If it's theta, as you had said earlier, then it's all right.

I've solved it, but the same problem remains. The accn a can't be expressed in terms of h'. If you do that, then the other things go away, but the time t is there. And anyway, expressing a in terms of h' seems very unphysical, since the accn a is constant.

You can try to solve it by almost the same method I'd outlined to you in post no. 3. But take T1 to be tension in the string connected to m1 and T2 for the other. All the other things are almost the same. This time, the torque on the pulley is (T1-T2)r. To eliminate T1 and T2, from the eqn 1 and 2, use this value of T1-T2.

Try it. I'm a bit too busy now. If you get stuck, I'll help you out. Also, I'll give a very neat method using energy conservation.

 

[Shooting Star]

I fail to see how the distance h' has anything to do with the acceleration.

Same here.

 

[Doc Al]

You can try to solve it by almost the same method I'd outlined to you in post no. 3. But take T1 to be tension in the string connected to m1 and T2 for the other. All the other things are almost the same. This time, the torque on the pulley is (T1-T2)r. To eliminate T1 and T2, from the eqn 1 and 2, use this value of T1-T2.

Done properly, he'll get the answer that he already gave in post #10.

I recommend that madchemist show the three equations (one for each mass and one for the cylinder) used to get the final answer. That will tell us if he solved it correctly or not.

 

[madchemist]

Okay, Torque=r(T1-T2) Nm=r[m1gsinb-um1gcosb-m1a-m2a-m2g] Nm= Mra/2 Nm. Solving for a gives: (2m1gsinb-2um1gcosb-2m1a-2m2a-2m2g)/M m/s^2= (m1gsinb-um1gcosb-m2g)/[(M/2)+m1+m2].

 

[madchemist]

I fail to see how the distance h' has anything to do with the acceleration.

That's what the title of my thread was all about. But I see now that it's not the distance h' by itself; rather it's the ratio that the vertical leg projection h makes with the bypotenuse distance h' to yield opp/hyp= sinb.

 

[Shooting Star]

Okay, Torque=r(T1-T2) Nm=r[m1gsinb-um1gcosb-m1a-m2a-m2g] Nm= Mra/2 Nm. Solving for a gives: (2m1gsinb-2um1gcosb-2m1a-2m2a-2m2g)/M m/s^2= (m1gsinb-um1gcosb-m2g)/[(M/2)+m1+m2].

Bravo. Now answer why did Doc Al have a hunch (or actually knew) that the answer in post 10 is going to be the same answer?

I'll show you another method soon. That's a bit faster.

 

[Shooting Star]

Using energy conservation

Frictional force F = uN1 = umgcos b. I’ll write y1 for dy/dt and y2 for d^2y/dt^2. It is the acceleration y2 which is to be found. Note that the linear speed v at the circumference of the pulley is y1 = rw, as before, where v=dy/dt.

KE of pulley = ½ Iw^2 = ½ (Mr^2/2)(y1/r)^2 = ¼ My1^2.

Suppose m1 has descended through a dist of y along the plane. Then, m2 has been raised by y, (what the OP called h’) and m1 has descended y/sin b vertically downward.

Decrease in PE = Increase in KE of the system + work done against friction =>
m1gy/sin b – m2gy = ½ m1y1^2 + ½ m2y1^2 + ¼ My1^2 + Fy =>
(m1g/sin b –m2g – F)y = (1/2 m1 + ½ m2 + ¼ M)*y1^2 =>
Ay = By1^2. (Writing A and B for the stuff in the brackets. We know what A and B are.)

Differentiating wrt t, we get,

Ay1 = B*2y1dy1/dt = 2By1y2 =>
y2 = A/2B, since the speed y1 is not zero. Now put in the values of A and B.

Quicker, isn’t it? ( I cannot vouch for small mistakes in algebra, though I’ve checked it twice.)