Please Help with this problem 6. 0.5/2 points | Previous Answers SerPSE9 15.AE.0

Question

Please Help with this problem

6. 0.5/2 points | Previous Answers SerPSE9 15.AE.006 Example 15.6 A Swinging Rod A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane (see the figure). Find the period of oscillation if the amplitude of the motion is small Pivot SOLVE IT Conceptualize Imagine a rod swinging back and forth when pivoted at one end. Try it with a meter stick or a scrap piece of wood CM Categorize Because the rod is not a point particle, we categorize it as a physical pendulum Analyze We found that the moment of inertia of a uniform rod about an axis through one end is (1/3)MI2. The distance d from the pivot to the center of mass of the rod is L/2. A rigid rod oscillating about a pivot through one end is a physical pendulum with d L/2 and, from 1 = (1/3Ma? Substitute these quantities into the equation: In one of the Moon landings, an astronaut walking on the Moon's surface had a belt hanging from his space suit, and the belt oscillated as a physical pendulum. A scientist on the Earth observed this motion on television and used it to estimate the free-fall acceleration on the Moon. How did the scientist make this calculation?

Explanation / Answer

time period T = 2*pi*sqrt(I/(m*g*L))

I = moment of inertia = (2/5)*m*r^2 + m*L^2

time period T = 2*pi*sqrt( ((2/5)*r^2+L^2))/(g*L))

T = 2*pi*sqrt( ((2/5)*0.02^2 + 0.366^2)/(9.8*0.366))

T = 1.215 s

--------------------------

part (b)

T' = 2*pi*sqrt(L/g) = 2*pi*sqrt(0.366/9.8) = 1.214 s

the period is 0.001 s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---