<p>So, post any questions or practice problems here.</p>
<p>What is the angular frquency of a physical pendulum. Derive the expression and explain the appx. required to treat it as SHM.</p>
<p>f = 1/T = 1/ (2<em>pi</em>sqrt(L/g)) = 2<em>pi</em>sqrt(g/L)</p>
<p>I think you can use the formula when the sin x = x, where x is measured in radians...this means that the angle has to be relatively small.</p>
<p>I think u did it wrong..........
Equation is something like f=1/T=1/(2pi<em>sqrt(I/(m</em>g*d)) , i dont know how to derive it though (tell us how..) and the condition for it to be considered SHM is that sixTheta is ALMOST equal to theta, and this is true for small values of theta measured in radians (~0-15Degrees)...</p>
<p>You're right...I made a mistake. What you saw in post 2 was the formula for a simple pendulum.</p>
<p>^^^ </p>
<p>indeed it is , and u can easily derive it from this formula too, knowing that the string is massless in a simple pendulum therefore the inertia of the system is that of a point mass, which is md^2, now the m and the d will cancel out with the ones under, and it will remain as L/g under the sqrt, same equation...</p>
<p>now we just need doh! to prove it (or u prove it if u know how)....</p>
<p>Let's make this thread an answer the physics c mechanics question posted above...and post (a) new question(s) (either MC or FRQ) for the next poster to solve. I did this for Chemistry last year...and it worked really well. (I find it hard to get myself to actually study...and this worked really great for me last year.) We just need people to participate and respond to a question within a day to keep this thread going.</p>
<p>Two questions...show all your work and reasoning</p>
<p>Questions 7 & 8 refer to a ball that is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equal to the asteroid's radius and then falls straight down toward the surface of the asteroid.</p>
<ol>
<li><p>What forces, if any, act on the ball while it is on the way up?
(A) Only a decreasing gravitational force that acts downward
(B) Only an increasing gravitational force that acts downward
(C) Only a constant gravitational force that acts downward
(D) Both a constant gravitational force that acts downward and a decreasing force that acts upward
(E) No forces act on the ball.</p></li>
<li><p>The acceleration of the ball at the top of its path is
(A) at its maximum value for the ball's flight
(B) equal to the acceleration at the surface of the asteroid
(C) equal to one half the acceleration at the surface of the asteroid
(D) equal to one fourth the acceleration at the surface of the asteroid
(E) zero</p></li>
</ol>
<p>I'm not very good at physics but I'll give this a try</p>
<p>8.D Fg=GM1M2/D^2 </p>
<p>distance is doubled, 2^2=4 so gravitational force is changed by 1/4. F=Ma, since M stays constant "a" must change by 1/4</p>
<p>7 A goes with number 8</p>
<p>I did 8 using a different style...
ma=Gmm/r^2
a = Gm/r^2
Gm/r^2 / GM/r^2 = r^2/r^2 = r^2/(2r)^2 = 4</p>
<p>But yeah...you did them both correctly. (Btw, you're supposed to post a question for the next poster to answer, but it's ok.)</p>
<p>A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension is 11,000 N. The speed vi of the elevator at the beginning of the 8 m descent is most nearly
(A) 4 m/s (B) 10 m/s (C) 13 m/s (D) 16 m/s (E) 21 m/s</p>
<p>Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass, as shown above. Each star has mass M and speed v. G is the universal gravitational constant. Which of the following is a correct relationship among these quantities?
(A) v^2 = GM/D (B) v^2 = GM/2D (C) v^2 = GM/D^2 (D) v^2 = MGD (E) v^2 = 2GM^2/D</p>
<p>bump .</p>
<p>Here are some more questions...most of which I don't know how to do, but I know the answers for:</p>
<p>4) IMAGE:<a href="http://img526.imageshack.us/img526/3397/15553198rj0.jpg%5B/url%5D">http://img526.imageshack.us/img526/3397/15553198rj0.jpg</a></p>
<p>9) IMAGE: <a href="http://img526.imageshack.us/img526/8011/91403669lf7.jpg%5B/url%5D">http://img526.imageshack.us/img526/8011/91403669lf7.jpg</a>
Two 0.60 kilogram objects are connected by a thread that passes over a light, frictionless pulley, as shown above. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60 kilogram objects as shown and the objects are released, the magnitude of the acceleration of the 0.30 kilogram object is most nearly
(A) 10.0 m/s^2 (B) 6.0 m/s^2 (C) 3.0 m/s^2 (D) 2.0 m/s^2 (E) 1.0 m/s^2</p>
<p>16) A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon. The person begins to climb the ladder at a uniform speed v relative to the ground. How does the balloon move relative to the ground?
(A) Up with speed v (B) Up with a speed less than v (C) Down with speed v (D) Down with a speed less than v (E) The balloon does not move.</p>
<p>24) When the block is set into oscillation with amplitude A, it passes through its equilibrium point with a speed v. In which of the following cases will the block, when oscillating with amplitude A, also have speed v when it passes through its equilibrium point?
I. The block is hung from only one of the two springs.
II. The block is hung from the same two springs, but the springs are connected in series rather than in parallel.
III. A 0.5 kilogram mass is attached to the block.
(A) None (B) III only (C) I and II only (D) II and III only (E) I, II, and III</p>
<p>25) A spring loaded gun can fire a projectile to a height h if it is fired straight up. If the same gun is pointed at an angle of 45° from the vertical, what maximum height can now be reached by the projectile?
(A) h/4 (B) h/(2sqrt(2)) (C) h/2 (D) h/sqrt(2) (E) h</p>
<p>31) IMAGE: <a href="http://img74.imageshack.us/img74/3136/31zi5.jpg%5B/url%5D">http://img74.imageshack.us/img74/3136/31zi5.jpg</a></p>
<p>33, 34, 35) IMAGE: <a href="http://img225.imageshack.us/img225/714/3335jn2.jpg%5B/url%5D">http://img225.imageshack.us/img225/714/3335jn2.jpg</a></p>
<p>If you can do any of these..and show me your work how you got these, I would be reeally really happy. I did a MC and these are the ones I can't figure out.</p>
<p>bumppppppp</p>
<p>I think 4 is B. You can limit it to B and D because it has to be to the left to cancel out the initial velocity and also partially up to make it go up. It has to be B because since the final velocity is less than the original so more work is done to stop the ball's horizontal movement than in creating the vertial movement.</p>
<p>I think 9 is D. The net force is ma or mg of the small block since everything else cancels out so it's 3N. This force acts on all 3 of the masses so using F=ma: 3N=(1.5)a, so a is 2m/s^2.</p>
<p>I think 16 is D. The center of mass of the ball-man system has to remain at the same point since no external force is acting. So, since the man is climbing up, he's raising the center of mass, so the ballon has to go down to offset that. It moves slower than v because it has a greater mass and an equal displacement on it's part affects the center of mass more than the man moving.</p>
<p>For 24...the picture would help, but im pretty sure it's none because all three of those change the elastic potential energy function of the spring. (i don't think elastic potential energy function is a real term but o well :))</p>
<p>post the answers</p>
<p>25 is a pain, but I got C. I'll call the original height h and the new one, H. You know the energy of the "shot" stays the same and was equal to mgh since it went that high before. So you can use mgh=1/2mv^2 to get the initial velocity of the ball when its shot...you should get: v=sqrt(2gh). Then you use the equation v^2=2<em>a</em>delta(x). a is g and x is the final height you want, H. So you should get that H=h/2.</p>
<p>nameless1...you got them all right. there is no picture for 24. 25 is a pain...i agree...i just did it...and it took me a while...b/c i was messing the v variable up.</p>
<p>here's what i did:
h = v^2/(2g) = v^2/20, v = the initial velocity (derived through KE=PE)
v = sqrt(2gh) = sqrt(20h), where h=height (derived through KE=PE)</p>
<p>the v of the thing shot at an angle = v of the thing shot straight up * sin(angle) = v*sin(45) = vsqrt(2)/2</p>
<p>now, h =v^2/20 = (v^2*.5)/20, keep in mind that v = the velocity of the thing shot straight up</p>
<p>h = (v^2<em>.5)/20 = (20h</em>.5)/20 = h/2...which is the answer of a lot of pain</p>
<p>31: not sure; the equation U=-Gm1m2/r leads me to believe its B or D...ill guess D
33: T=2<em>pi</em>sqrt(l/g) so make l=.25 A
34: force = mg(3/5)+mg(4/5)*.3= 42 (mg(4/5) = normal force) C
35: you know: M2l1=ml2 and ml1=M1l2 so solve the first one for m to get m=M2l1/l2 and then use the second equation to get l1/l2=M1/m and plug that in to get m=M2M1/m which simplifies to m=sqrt(M1M2) E?</p>
<ol>
<li>D</li>
<li>A</li>
<li>B</li>
<li>E</li>
</ol>
<p>I'll explain 34:
F = sideways component of weight + Fk
F = mgsin(x) + umgcos(x)</p>
<p>then u substitute in and get 42 N</p>