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Need help solving homework problems

  1. Jul 22, 2005 #1
    Please help i'm totally lost an i don't understand how to do these problems.
    Could u please explain these problems out so i can better understand them and solve too.I am failing physics right now that's how clueless i am about this subject.

    1. Consider two uniform solid spheres where both have the same diameter, but on has twice the mass of the other. The ration of the larger moment of inertia to that of the smaller moment of inertia is?

    2. A force is applied to the end of a 2 ft long uniform board weighing 50 lb, in order to keep it horizontal, while it pushes against a wall at the left. If the angle the force makes with the board is 30 degrees in the north western direction, the magnitude of the applied force F is?

    3. A solid ball, solid cylinder and a hollow pipe all have equal masses and radii. All three are released simultaneously at the top of an inclined plane, which will reach the bottom first?

    4. A hoop of radius 0.5 m and a mass of 0.2 kg is released from rest 5 meters above the horizontal and allowed to roll down an inclined plane. How fast is it moving when it is 2 meters above the horizontal?

    5. A wheel with a rotational inertia of 8.0 kg-m ^2 and a rotational kinetic energy of 155 Joules requires what torque to bring it to rest in 4.0 s?

    6. A diver can change her rate of rotation in the air by "tucking" her head in and bending her knees. Let's assume that when she is stretched out straight she is rotationg at 1 revolution per second. Now she goes into the "tuck and bend," effectively shortening the length of her body by half. What will her rate of rotation be now?

    Please help i am very clueless on where to start of how to finish these problems. I don't know anything about physics.
     
    Last edited: Jul 23, 2005
  2. jcsd
  3. Jul 22, 2005 #2

    amt

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    For 1.
    You need to find the formula for the Moment of Inertia of a Sphere. I think it's (2/5)MR^2, but you might want to check. Now the problems says that one Mass is twice the other. Simple, make m1=m and m2=2m. Substitute this in the formula for Moment of Inertia and devide one by the other.

    For 2.
    Looks like vector addition. Did you try using F=ma? Since it contains angles, your problem will contain trignometric identities.
     
  4. Jul 23, 2005 #3
    Calculate the time for one of the objects and see how the value changes if you double the mass/radius/rotational inertia.

    You can use conservation of mechanical energy on this one.
    [tex] E_H = E_{H- \Delta h} + \frac{1}{2}J \omega^2 + \frac{1}{2}m v^2 [/tex].
     
  5. Jul 23, 2005 #4
    That's the correct expression :smile:.
     
  6. Jul 23, 2005 #5
    thank u to all replied still kinda confused but thanks 4 ur help anyway
     
  7. Jul 23, 2005 #6

    amt

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    For 1.
    It says devide the larger one by the smaller one(ratio of the large one to the small one). The larger one is the one with mass = 2m. The smaller one is simply m.

    So-

    [(2/5)2MR^2]
    ------------- = 2
    [2/5)MR^2]
     
    Last edited: Jul 23, 2005
  8. Jul 23, 2005 #7

    amt

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    For 2.
    Here is the free body diagram. Does it look correct?


    |wall
    |
    |<-----2 ft board------>
    |==================|\
    |..............|................ |.\
    |..............|50 lbs.........|30\
    |..............|.................|....\
    |..............V.................|.....\ force pushing up is 30 degrees.



    Pushing is done to prevent to board from falling.
     
  9. Jul 23, 2005 #8
    somewhat

    the 30 degree angle goes on top from the right end to the wall.
    How do i figure that out with the equation f=ma when i don't have the acceleration.
     
  10. Jul 23, 2005 #9
    The system is in equilibrium, what does that tell you about the aceleration?
     
  11. Jul 23, 2005 #10

    amt

    User Avatar


    OK, I have still not figured out this yet. But since you have requested some teaching steps, here is one fundamental approach you absolutely have to know. This will help you with most physics problems.

    See the pushing force at 30 deg? Since this force is at angle it can be broken into 2 components. One is Fx and the other Fy as shown here

    ..........|\
    ..........|.\
    Fy=50..|..\..We need to know this Hypotonuse length which is the reqd force.
    ..........|...\
    ..........|....\
    ..........v----\ Fx=? We have to find this as well.

    So basically we have Fy which is 50. Now we need to find Fx and Fr the required force(the 50 might be a negative value since it is opposite to the push)

    Now you have to use simple trig formulas:

    Sin@ = opposite/adjacent.
    Cos@=adjacent/hypotonuse etc etc.
     
  12. Jul 23, 2005 #11
    thank u that helped alot
     
  13. Jul 23, 2005 #12
    need someone to explain please

    could someone please explain this equation for me
    [tex] E_H = E_{H- \Delta h} + \frac{1} {2} J \omega^2 + \frac{1} {2}m v^2[/tex]

    this is the equation just incase the one above didn't come out right
    EH= EH- Delta h + 1/2 j omega^2 + 1/2 mv^2
     
  14. Jul 23, 2005 #13

    amt

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    I will try to post something for problem-2. It's been awhile since I opened my Physics books. Perhaps someone else here might also offer some insight.
     
  15. Jul 23, 2005 #14

    amt

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    Looks like E is the energy. 'omega' should be angular velocity.
    (1/2)mv^2 is the kinetic energy.
     
  16. Jul 23, 2005 #15

    amt

    User Avatar

    OK since none of the Physics gurus have responded, I am going to write my own calculations. But no guarantee this is correct. However, the answer looks good and reasonable. You should atleast get partial credit for this. Try to fully understand how the problem is broken into.

    |..............N
    |..............^
    |..............|
    |..............|
    |..............|
    |1xN...........................Fy
    |<=================^\
    |..............|................ |.\
    |..............|222N..........|30\
    |..............|.................|....\
    |..............V.................|.....\
    |.................................|......\
    |____________________|____60\ ----->force Fx

    Understanding the problem.
    The 2 feet board length is not needed. It might have been given to throw you off. Also note that the Co-efficient of friction(u) is not given. The board is being pushed against the wall. But it will not move past the wall. So we can safely assume that 'u' is = 1 (which is the maximum coefficient of friction possible). Think about it...the more and more 'u' is, the harder it gets to move anything, so since you can't move it past the wall, we can say it's the max:1. So the force acting on the wall is the coefficient of friction times the Normal force(N) acting up or Fk=1xN.

    Now the weight of the board is 50 lbs and is pushing down. We first must convert this to Netwons(N). 1 lb = 4.44N, so 50 lbs = 222N. This is the new force pushing down.
    Since this force is pushing down, there is an equal and opposite force pushing up. This is known as N, commonly known as the Normal force(see diagram).

    So now you recognize 4 quantities. You know the force acting down (222Newt) and the force acting up(N), and the coefficient of friction (u=1) or more specifically Fk=1xN.

    We will now call the force pushing the board at 30deg as 'T' (in physics books it's referred to as tension). T is the hypotonuse of the triangle.

    Basically we want to get 4 forces into an equation. The up(N), down(weight), left and right.

    Since T is at angle, it has to be broken into 2 components. We will call this Fx and Fy. Fx acts horizontally and Fy acts vertically. OK?

    Fx makes the base of the triangle, Fy makes the height of the triangle and T makes the hypotonuse of the triangle.

    Look at the diagram. I have a new angle of 60 degrees. The reason I put that there was, it would be easy to work with Sin and Cos instead of using other functions. You should know that all 3 angles should add to 180deg.

    Breaking down T into x and y components
    Cos(60) = adj/hyp = Fx/T or T.cos(60) = Fx ------(1)This is the horizontal component of T.
    Sin(60)=opp/hyp=Fy/T or T.Sin(60)=Fy-----(2) this is the vertical component of T.

    Creating the equations
    Let's take the horizontal forces. From now on we don't have to worry about the angle 30 with T because they have been replaced with x and y components.

    Taking equation-1.

    T.Cos(60)- 1(N) = 0.............(3)
    T.cosn(60) is the force acting to the right and 1.N is the coeff of friction times the Normal force which we simply know as N. Simply stated, the force acting on the right is counter acted by a force on the left. So we put them together and subtract them because they are in different directions.

    Taking equation-2.
    T.Sin(60) + N - 222 = 0.........(4)

    Here We take all the vertical forces and balance them just like we did in the previous step. N is the normal force acting straight up to balance the weight and 222 is the weight in newtons pushing down (so it's negative). Fy is the vertical component we got from T.

    We can now solve equation-4 for N.

    N=222-T(sin60)......(5)
    We can now put this in equation-3 to get:
    T.cos(60)-(222-T.sing(60)=0
    =>0.5T-222+0.866T=0
    So, T=162 Newtons which is the force which was asked in the problem.


    You can take it a step further by finding the normal force N by using equation-5 again: N=81.7N

    ========================================================
    The answer seems correct. Think about it, if you are carrying something weighing 222N and if you get tired, you lean it against the wall. The wall shares some of the load to give you some relief. So, 162N would be the force you are carrying and 60N would be the weight carried by the wall(162+60 = 222N) if you were leaning at 30deg. This ratio can be varied by changing the angle with which you are leaning it against.
    I hope this is correct. If not someone please correct me.
     
    Last edited: Jul 23, 2005
  17. Jul 24, 2005 #16
    4. A hoop of radius 0.5 m and a mass of 0.2 kg is released from rest 5 meters above the horizontal and allowed to roll down an inclined plane. How fast is it moving when it is 2 meters above the horizontal?


    Let [tex] h_0 = 5m [/tex], [tex] \Delta h = 3m [/tex], [tex] m = 0.2kg [/tex] and [tex] R = 0.5m [/tex]. Since it's a hoop we get [tex] J = mr^2 [/tex].
    We know that [tex] \omega = \frac{v}{R} [/tex].

    Let's use conservation of mechanical energy:

    Potential energy in the beginning = potential + kinetic + rotational energy in the end.
    [tex]E = mgh_0 = mg(h_0 - \Delta h) + \frac{1}{2}mv^2 + \frac{1}{2}J \omega^2 [/tex]
    That is [tex]gh_0 = g(h_0 - \Delta h) + \frac{1}{2}v^2 + \frac{1}{2}R v [/tex] and finally [tex] 2g \Delta h = v^2 + Rv [/tex], by solving this we get [tex] v [/tex] and [tex] \omega [/tex], you can solve this by your self, can't you? :smile:
     
    Last edited: Jul 24, 2005
  18. Jul 25, 2005 #17
    I just want to say thank u to everybody that helped me and a special thanks to amt u helped me alot thanks
     
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