TWJian Posted March 5, 2008 Share Posted March 5, 2008 (Please delete the other duplicate thread. Sorry. Refreshed by accident) Recently I came across a confusingly-worded question (at least to me) in Physics. Semantics might matter, so I'm putting the entire question down ad verbatim: Four particles, each of mass 0.20 kg, are placed at the vertices of a square with sides of length 0.5m. The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis A that passes through one of the particles. The body is released from rest with rod AB horizontal. Find: a) the rotational inertia of the body about axis A b) the angular speed of the body about axis A when rod AB swings through the vertical position I'm assuming that the square is arranged like this, on an xz plane: X X A B With the rotational axis at A and pointing out(or in, doesn't matter( of the page. The square would then rotate like a wheel. and the CoM would have a mass of 0.8kg at the center of the square. To find I, let I=I(CoM) +mh^2 (Parallel axis theorem, where h is the perpendicular distance between the axis through the center of mass (CoM) and A) h is therefore squareroot(2*0.25^2) right? Now how do I calculate I(CoM)? Assume it's a square and use I(com)=1/12*m(a^2+b^2), with a=b and I(com)=1/12*0.80(2x0.25^2)? Doesn't seem right as the mass are concentrated at the four vertices instead of being evenly distributed throughout the square...So... Assume it's a hoop with negligible thickness since the movement of the vertices appear to trace a circle? (am I even right?) Use the equation I(com)=mR^2, with r=h (from mh^2) and I(com)=0.80 squareroot(2*0.25^2) ? This interpretation(of the rotation) is even worse for (b). I'm assuming that energy is conserved and therefore: [mgh + 0.5*I(angular velocity)^2]initial= [mgh+0.5*I(angular velocity)^2]final Since AB is horizontal and ends up as vertical, There is no net change of h and therefore the gravitational potential energy is the same on both sides. With angular velocity(initial)=0 (since the square is released from rest), wouldn't angular velocity(final) be zero as well? That is, assuming the square and even rotate around A without any provided torque in the first place, since the CoM would definitely have to move upwards first before moving down to it's original position, thereby nessacating an input of energy due to increased mgh. So, how does the blasted square even move (except by falling down in translation motion if A is not locked down) in the first place? Obviously I am wrong. But where? How should I answer this question properly? Thanks for any help. Link to comment Share on other sites More sharing options...

swansont Posted March 5, 2008 Share Posted March 5, 2008 The rotation isn't about the CoM, so you needn't worry about that. All rotation is about A. What are the moments of inertia for all of the particles for rotation about A? Link to comment Share on other sites More sharing options...

TWJian Posted March 6, 2008 Author Share Posted March 6, 2008 Got it. Thanks. Apparently I was doing too many parallel axis questions and was mind was swimming with them. Managed to find three ways to solve (a), though two of them are quite similar. The obvious solution 9and apparently shortest) is I(A)= 0.2kg (0+0.5^2+0.5^2+2*0.5^2)m^2 = 0.2 kg*m^2 As hinted by you. 2) Let d be the distance between two vertices. Since the vertices of the square trace a circle with radius, r= squareroot(2*(d/2)^2), use the formula: I(A)= I(CoM)+mh^2 =mr^2+ mh^2, h=r I(A) = 2mr^2=0.8kg*2*2*(0.25^2)m^2 =0.2 kg*m^2 3) Similiar to above, but proves that I(CoM) of a square with point masses at vertices is similiar to a circular band of negligible thickness. I(A)= I(CoM)+mh^2 = 4m(particle)*(distance of particle from CoM)^2+mh^2 = 4*0.2kg*(2*0.25^2)m^2 + 0.8kg(2*0.25^3)m^2 Note: same as above = 0.2kg*m^2 Therefore, the rotational inertia of a square with the only point masses at the vertices over an axis passing through one of the axis perpendicular to the plane the square is drawn in is: I= Total mass*Length of square^2 And for rotation with the axis passing through the CoM: I=Total mass*Length of square^2*(1/2) For a circular band rotating with an axis passing through it's fringes (or a particle at distance r from the center): I=2mr^2 Now for (b) Apparently I thought the blasted square has to rotate counter-clockwise despite being able to rotate clockwise... D C A B to A D B C change of height, delta h=h(i)-h(f)=0.5m (Initial gravitational potential energy)+(Initial rotational kinetic energy)= (Final gravitional potential energy)+(Final rotational kinetic energy) Since initial rotational velocity=0, mgh(i)=mgh(f)+0.5I(rotational velocity)^2 rotational speed=squareroot[2mg(delta height)/I] (speed is positive since it is the magnitude of velocity (in this case)), despite velocity being negative due to clockwise rotation. Link to comment Share on other sites More sharing options...

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