repost, i am in college physics for algebra and while i wish i was in calculus a

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repost, i am in college physics for algebra and while i wish i was in calculus and understand it i need to know the algebraic way ... also i have 2 different approaches to part a that are giving two different answers... and i dont need like all the work but please at least like an explanation good enough for me to put the work ionto and understand

i did T= 2*pi*(L/g)^.5

so Lrock is .5 : is T/5

anyways solving i either get 1.12m or .02 m for part a... i think its because in my algebra ... when i cancel the g constant out i get the larger value! (1.12m)

so i mean any help appreciated .. using help and .02 m for part b i got 24 * (the mass of the girl not given) so i want to make sure thats right

part c is well again im not supposed to use calculus but someone said we can take intigral of I'=.5l and i get that i guess ut i need to get it algebraicly

thanks guys

This week we started toexplore oscillations. We closely observed the oscillation of a spring in the classroom and in the lab. We also looked at a pendulum in class. You might have seen a nice grandfather’s clock as an application of the physics of a pendulum, here is another real world application:An inquiring studentthat has a mass of 70. kgis watching a wreckingball that is attached to a cranedemolish a buildingnext door. The wrecking ball is swinging parallel to the property fence between her house and the neighbor’s.By running some experiments the student is able to estimatethe length and mass of the wrecking ball aswell as predict itsmotion.

A.In an effort to estimate the length of the cable on the wrecking ball the student fashions a pendulum from a rock and a string. Her pendulum turns out to be0.5m long. Shethenobserves that the wrecking ballcompletesone oscillation in the same amount of time that it takes the rock to complete 5 cycles. What is the length of the cable on the wrecking ball?

B.The wrecking ball is swinging at just the right height so that the student can stand at the bottom of the trajectory(lowest point)and grab onto the ball to swing –something that is just way too tempting for this inquisitive student. In order to justify invading the construction zone, before the student jumps on sheattaches a magnet to an accelerometer and sticks the accelerometer onto the iron ball. The maximum velocity reading of the ball before she jumps on is1.98m/s. Once she is on the ball, she reads that the maximum speed of herself and the ball together is1.90m/s. Whatis the mass of the wrecking ball?

C.While swinging on the wrecking ball, the student notices that, at first, the wreckingball is in line with her birdbath when it is at a peak in its trajectory, and in line with her front door at the lowest point in the trajectory. However, since drag is acting against the wrecking ball’s motion, 3 minutes later the peak of the wrecking ball’s trajectory is in line with her garden gnome, the halfway point between her birdbath and front door. How much longer will it take for the peak of the ball’s trajectory to be halfway between her garden gnome and her front door? From the time she begins to watch the wrecking ball, how long will it take for the trajectory peak to be in line with the edge of her welcome mat, which is one tenth of the distance from her front door to the bird bath?Thereis a picture on page 3 that that might help you visualize thisproblem.

Explanation / Answer

A) time period of rock is T/5 where T is time period of wrecking ball.

T=5t

2pi (L/g)0.5=*2pi(l/g)0.5

cancelling pi, then squaring both sides, then cancelling g,

L=52*l=25*0.5=12.5 m

B) This will be solved on the concept of balancing angular momentum, Let the mass of ball is M, mass of the girl is given 70kg

angular momentum before she is on the ball= angular momentum when she joins the ball

M*1.98 L= (M+70)*1.90*L...............................(the formula of angular momentum is mass *velocity*length)

1.98M=1.90M+70*1.9

0.08M=70*1.9

M=70*1.9/0.08= 1662.5 kg

C) It is decaying harmonic motion as its amplitude is decreasing, with amplitude having half life of 3 minutes

Its amplitude becomes 0.5A after 3 min, again it becomes half after another 3 minutes and so on.

0.5A..........3min

0.25A.........6 min

0.125A....9 min

0.0625A...12min.....and so on..

since it 3min longer for the peak of the ball’s trajectory to be halfway between her garden gnome and her front door.

distance of welcome post= 0.1A......clearly it takes more than 9 min but less than 12 min

Let the number of such 3 min cycles be x

0.5x=0.1

to find x, take log on both sides,

x*log0.5=log0.1

x=log0.1/log0.5=3.32 (u can calculate this on google or just calculator)

total time =3min*3.32=9.97 min= nearly 10 min

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