Hi,
I found this interesting thread,
https://www.physicsforums.com/threads/accelerating-a-car-including-the-moment-of-inertia-of-the-wheels.930374/
but as it has been closed to replies, I decided to ask here.
The thread ended up with the equation:
where
τ - 200Nm engine torque provided on...
For this problem,
For part(a) the solution is,
However, how did they know that the max speed is reached in the vertical position?
For part (d) the solution is,
However, I thought the solution would be because there is net gravitational torque in clockwise direction (torque due to small mass...
The block starts to slide if friction can no longer hold the block.
F=u*n and F=(m1+m2)a
so: (m1+m2)a=uN=>am1+am2=uN=>am2=(uN)/(am1)
So:am2=(uN)/(am1) is the force.
The answer is F=(u*m1g(m1+m2))/m2
I do not see how the acceleration terms are canceled. Is my answer equivalent to this?
Hi,
I am not sure if I have derived the matrix correctly, because of my results in task b
I solved task 1 as follows, I assumed that all three particles move to the right
$$m \dot{x_1}=-k(x_1 - x_2)$$
$$2m \dot{x_2}=-k(x_2-x_2)-3k(x_2-x_3)$$
$$3m \dot{x_3}=-3k(x_3-x_2)$$
Then I simply...
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How is the mass inertia product calculated? I have two examples and each one uses something different.
Example 1:
Example 2: moments and product of inertia of the cylinder
my solution:
a)
F(upward)=Fb +Fw
=(1.3+1.5)X9.8
=27.44N
total Mass = 2.8kg
b)
Volume increased = π(0.2/2)^2 x 1.5/100
=4.7x10-4 m^3
T+Fb =mg
T=mg-Fb
T=2.2x9.8 -1000 x 4.7x10-4 x 9.8
T=17.4N
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where ##v = \dot{x}##, where x is my coordinate and f is a function of v only (no explicit dependence on t or x). What I get by solving the Euler-Lagrange equations is:
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I tried approaching this question like this:
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and theta_dot = v/R since R is constant
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I can write the 3.5 kg block equation as Fnet(block 1)=(Force of tension)-(Force of friction)=m1a
I can write the 2.8 kg block as Fnet(block 2)=(Force of tension)-(Force of gravity2)=m2a
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tried writing the x position as
x = Acos(wt) (ignoring the phase)
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Substituting that into the individual motion equations would get the required result for the individual masses, but I am not sure how to combine the equations to get the reduced mass
I got 13N but is that right because apparently, it's wrong
Here's my work:
F = mg = 2(10) = 20N
F = ma
a = F/m = 20/4+2 = 20/6 = 10/3 = 3.3m/s^2
T = mg - ma T = (2kg)(10m/s^2) - (2kg)(3.3m/s^2) = 13.4 N
I appreciate it! And if I'm wrong could you show how you got your answer? Thanks
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