i have the physics\' review sheet for the final exam, i have got most of answers

Question

i have the physics' review sheet for the final exam,

i have got most of answers but the very last problems, 13, 14, 15

most of them are about the torque, and inertia i think...


well i tried and i want to check my answers, so could anyone help me with those three problems?

if you dont have enough time to do all three,

you can just do a one problem. i will still very appreciate.


thank you!!




i took picture of the problems, since the problems contain diagrams.

if the image down there is too huge, you can save the image by right clicking your mouse in your computer and work on it :)






ps.

some of them i circled two answers simultaneously, since i was not really sure...

for #13

Find the final angular velocity ?f of the bullet and rod system:

Conservation of angular momentum equation:

(1) Lf = Li,

where Lf is final, Li is initial.

NOTE: I'll use plain L (no subscript) for the length of the rod.

Since the rod is initially at rest, we need only consider the Li for the bullet:

(2) Li = m * Vv X r,

where:

Vv = the velocity vector
r = the radius of the circle of rotation

Substituting:

(2') Li = (M / 9) * v * sin30 * (L / 2)

= M * v * L / 36


The final angular momentum of the combined bullet and rod about the CM:

(3) Lf = If * ?f,

where:

(4) If = sum of moments of bullet and rod

= (M / 9) * (L / 2)^2 + M * L^2 / 12 (from Source 1)

= M * L^2 / 9


Substituting (2') and (3) into (1):

(5) If * ?f = M * v * L / 36

and solving for ?f:

(6) ?f = (M * v * L / 36) / If

and substituting (4) for If:

= (M * v * L / 36) / (M * L^2 / 9)

= v / 4L, which is (b)

but i am not sure my answer and set up is correct... can you double check it for me??

for problem 14, i pretty much guessed...

for problem 15,

the angle is 30, due to the R (hypotenuse) is the double of h, i used pythegorean theorm, to find the angle.

balancing the torque,

F*(R-h)=mgRCos(30)

F=[mgRCos(30)]/(R-h)=?3mg, so answer is (e)


but i dont understand that, lets say if the step is higher, so angle is 45

then it would be

F=[mgRCos(45)]/(R-h) = ?2mg ??

which means, bigger angle, but less force to push the bicycle wheel????????

i dont understand!

Explanation / Answer

I will do the 15th for which I have got answer as e) You have done simple errors. My advice: You seem to be thinking too much. Its all simple, just have the concepts strong. Balancing torques... From the geometry angle between horizontal and line from center of the ball to contact edge is 30 degrees, and sin30=0.5 F*R*sin30=(m*g)*R*cos30 R cancels out. F=m*g/tan(30) or F=m*g*sqrt(3) Hope that helped :)

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