The rigid body shown in Figure 10-66 (it is a triangle with the upper mass = to M and the bottom two corners/masses equal to 2M. The bottom side length equals .70 cm with point P in the middle of the side and the other two sides equal .55 cm)consists of three particles connected by massless rods. It is to be rotated about an axis perpendicular to its plane through point P. If M = 0.40 kg, a = 35 cm, and b = 55 cm, how much work is required to take the body from rest to an angular speed of 5.0 rad/s?
*My work so far*
I believe I need to use:
W = (1/2)(I)(ang speed)^2
I am not sure how to find I though. Would I use I = I1 + I2 + I3 and calculate I for each of the 3 masses?
For Further Reading
* Physics - KE/Inertia - bobpursley, Monday, March 26, 2007 at 4:03pm
Yes, use the addition formula to get the moment of inertia. In each case, the distance is the distance of the mass from point P.
My work:
I found I1 + I2 + I3 = Itotal
I1 = (.80)(.35)^2 = 0.098 kg*m^2
I2 = (.80)(.35)^2 = 0.098 kg*m^2
I3 = (.40)(.18)^2 = 0.01296 kg*m^2, for this r I found h of the isosceles triangle by doing h=sqrt(b^2-.25(a^2))
I added up the 3 I's and got Itotal to be 0.012946 kg*m^2.
Next I used the equation:
Work = .otal*omega^2
Work = .5(0.012946)(5.0)^2 = 2.612 Joules
This is apparently the wrong answer but I don't know why. Do you see any mistakes? Thanks!
Never mind...I forgot to take the square root when finding h. I got it now.
To find the moment of inertia (I) of the rigid body, you are correct in using the formula I = I1 + I2 + I3, where I1, I2, and I3 are the moments of inertia for each of the masses.
To calculate I1, you need to find the distance of mass M from point P. In this case, the distance is a/2 because point P is in the middle of the side of length a. Given that a = 35 cm, I1 = M * (a/2)^2 = 0.40 kg * (0.35/2)^2.
To calculate I2, you need to find the distance of the two masses 2M from point P. Since these two masses are at the corners of the triangle, their distance from point P is b/2. Given that b = 55 cm, I2 = 2M * (b/2)^2 = 2 * 0.40 kg * (0.55/2)^2.
Lastly, to calculate I3, you need to find the distance of mass M from point P. In this case, the distance is the height of the isosceles triangle, which can be found using the formula h = sqrt(b^2 - 0.25*a^2). Given that a = 35 cm and b = 55 cm, I3 = M * h^2 = 0.40 kg * (sqrt(55^2 - 0.25*35^2))^2.
After calculating I1, I2, and I3, you can add them up to get the total moment of inertia (Itotal). From there, you can use the equation Work = 0.5 * Itotal * (angular speed)^2 to calculate the work required to take the body from rest to an angular speed of 5.0 rad/s.
Glad to hear that you were able to solve the problem. Taking the square root when finding the height of the isosceles triangle is indeed necessary to obtain the correct value for I3. It seems like you have correctly calculated the moment of inertia for each mass and then added them up to find the total moment of inertia (Itotal).
After finding Itotal, you can use the equation W = (1/2)(Itotal)(ω^2) to calculate the work needed to take the body from rest to an angular speed of 5.0 rad/s. Plug in the values you have obtained:
W = (1/2)(0.012946 kg*m^2)(5.0 rad/s)^2
W = 0.01618225 J
So the work required to achieve the desired angular speed is approximately 0.0162 J.