5 Steps to a 5 by Greg Jacobs

5 Steps to a 5 by Greg Jacobs

Author:Greg Jacobs
Language: eng
Format: epub
Publisher: McGraw-Hill Education
Published: 2018-08-06T16:00:00+00:00


4. D—The angular position function is given by the integral of the angular velocity function with respect to time. The limits on the integral are 2 and 3 seconds:

this evaluates to approximately 27 radians. Using x = rθ, the distance traveled is closest to 14 m.

5. (a) The rotational inertia of the entire contraption is the sum of the moments of inertia for each part. I for the rod is given; I for a point mass a distance L from the pivot is mL2. So, Itotal = (1/3)ML2 + mL2. Be sure to differentiate between M and m.

(b) Rotational kinematics won’t work here because angular acceleration isn’t constant. We must use energy.

U1 + E1 = U2 + K2

Define U = 0 at the bottom of the contraption when it hangs vertically. Then, U2 is only caused by the rod’s mass, which is concentrated L/2 above the zero point, so U2 = MgL/2. U1 is due to all of the mass, concentrated L above the zero point: U1 = (M + m)gL. K1 = 0, and K2 is unknown.

(M + m)gL + 0 = MgL/2 + ½ Iω)2.

Plug in I from part (a) and solve for ω to get



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