An experiment is performed with thin, uniform metal rods of...

An experiment is performed with thin, uniform metal rods of different lengths, each pivoted around a point one third the length of the rod from the top end and used as a physical pendulum, as shown in the figure on the previous page. Each rod is pulled back so that it makes a small angle with the vertical and is then released, and the period of oscillation is measured. The data are recorded in the table below. egin{tabular}{|c|c|c|c|c|c|} hline Length (m) & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 \ hline Period (s) & 1.2 & 1.6 & 2.0 & 2.3 & 2.6 \ hline & & & & & \ hline & & & & & \ hline end{tabular} (c) Indicate below which quantities should be graphed to yield a straight line whose slope could be used to calculate a numerical value for the acceleration due to gravity g. Horizontal axis: Vertical axis: Use the remaining rows in the table above, as needed, to record any quantities that you indicated that are not given. (d) Plot the straight line data points on the graph below. Clearly scale and label all axes, including units if appropriate. Draw a straight line that best represents the data. (e) Using your straight line, determine an experimental value for g. Free Response 2 (15 points) A uniform rod of mass M and length L is pivoted at a point a distance L/3 from the top end, as shown above. The rod is pulled back so that it makes a small angle with the vertical and is then released. (a) Using integral calculus, show that the rotational inertia for the rod around its pivot is ML²/9. I = ?r²dm = ?x² (M/L dx) = M/L ?_{-2L/3}^{L/3} x² dx = M/L ( (L/3)³/3 - (-2L/3)³/3 ) = M/L ( L³/81 + 8L³/81 ) = M/L ( 9L³/81 ) = ML²/9 (b) i. Using Newton's 2nd law in rotational form, write but do NOT solve a differential equation in terms of M, L, and physical constants, as appropriate, that can be used to determine the angular displacement theta as a function of time t. ii. Using the differential equation from part (b)i, show that the period of oscillation for the rod is given by the expression T = 2??(2L/3g).

Transcript

00:02 So this is a free response question involving a rotating object and we're going to first calculate the rotational inertia, which has already been done, and then we're going to use newton's second law for rotation to derive an equation which can then be verified through an experiment and the experiment will be used to derive a value for g.

00:30 So the first part has already been done.

00:35 We just took the formula for the rotational inertia is just equal to the integral over the distance from the x rotation squared, which is called r squared, times a differential piece of the mass, and then correct the reason here was that dm is going to be equal to the linear mass density, which is m over l, since it's a uniform rod, times dx.

01:13 And so that was rapidly converted to this is just equal to x squared m over l times dx.

01:22 And the only thing that then had to be done was to determine the limits of integration here.

01:29 Since we are calculating the rotational inertia about a point that is one third of the way down the bar, we start with our x being a negative two thirds the way down the bar and going up to a positive one third of the way up the bar from the axis.

01:50 And then after carrying out the calculations, it does indeed turn out that the rotational inertia of the rod about the pivot is ml squared over nine.

01:59 So that part was done.

02:02 Now the second part is it even tells us to use newton's second law in rotational form to find a difference equation.

02:09 Newton's second law is often how we have to find a difference equation because it does give a derivative of something, the acceleration being the second derivative of the position.

02:25 And what we're going to do here is since they told us to find, use newton's second law in rotational form, we need to consider the torque, because newton's second law for rotation says that the torque is equal to the rotational inertia, which we just found out what it is now, times the rotational acceleration.

02:46 Well, if we consider the forces acting on this bar, there's a force acting at the pivot point there.

02:57 But if we make the pivot point our axis of rotation, there is no torque exerted by the forces acting at the pivot point, because that is at the axis.

03:07 So the only force acting on our bar is the force of gravity, which we can treat as acting at the center of the bar.

03:17 And the magnitude of the force of gravity is going to equal m times g.

03:22 And it's going to be acting straight down.

03:25 So the angle that it makes with the position of the bar is the angle theta that they gave us right here.

03:36 And so from the definition of torque, we have that the torque is equal to the force, which is m times g.

03:49 And the distance that the force is from the axis of rotation, which in this case is going to be, well, the center of mass of the object is at l over 2.

04:08 But then i'm going to subtract from that, from the end, it's l over 2 from the end, but we would need to subtract from that the l over 3, which is the position of the axis of rotation.

04:26 And so the distance of the center of mass from the axis of rotation is, well, halfway down the bar.

04:36 That gets you from the end of the bar to the middle of the bar, which is where the force of gravity is acting, minus the distance from the end of the bar to the axis of rotation, which is l over 3.

04:50 And so that is going to be the distance from the axis of rotation to the center of mass.

04:56 This turns out to be l over 6 then.

04:59 And that is the force, and that times our distance to our axis of rotation.

05:05 Now we need to consider that since it is a torque we're trying to calculate, it's going to be the sine of the angle between our line of action of our force and the line from the axis of rotation to where the force is being applied.

05:21 But that is the angle that they have given us here.

05:25 So we have everything we need for our torque.

05:28 One more thing though, this is a restoring torque.

05:32 If we swing the bar to the left, then the torque is going to be to the right, and if we swing the bar to the right, the torque is going to be left.

05:40 So we would need to put a minus sign in front of this to give us the direction that the torque is acting.

05:48 Now that, according to newton's second law of rotation, must equal i times alpha.

05:59 Let's clean this up just a little bit.

06:04 L over 2 is 3l over 6, and l over 3 is 2l over 6, so that leaves l over 6.

06:15 So this becomes a minus mg times l over 6 times the sine of theta is equal to i, and i we figured out is equal to ml squared divided by 9 times alpha.

06:38 Now we have an m on both sides, so we can divide both sides of our equation by m, and we have an l on one side and l squared on the other side, so we can divide by l and get rid of at least one of the l's over on the right and the l on the left.

06:55 And so that leaves us with saying a negative g over 6 times the sine of theta is equal to l over 9 times alpha, where alpha is the angular acceleration.

07:23 We can clean this up just a little bit more, get everything over on the left -hand side that we can.

07:32 So this is a minus 9 over 6, multiplying both sides by 9, times the g, and we can divide both sides by l and get that over here, so that's 9 times 6 times g over l times the sine of theta is equal to the angular acceleration.

07:56 The angular acceleration is the second derivative of the angular position theta with respect to time, and so now we have this, and we can even simplify it even further, 9 times 6 becomes 3 halves, so this is 3 halves, which is right on both top and bottom by 3, times g over l times the sine of theta is equal to the second derivative of theta with respect to time.

08:37 This is a differential equation that could be used to determine the angular displacement theta as a function of time t...

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