物理讨论专区

Lov瑜瑜4ever

U mean the power de ques?? I got!
vick5821 发表于 5-8-2010 08:04 PM

不是拉
是417楼的那个问题。。。

cheesit92

回复 417# vick5821



用principle of conservation of angular momentum,
L2=L1
I2w2=I1w1   , I=mr^2
w2m(r2)^2=w1m(r1)^2
w2=(r1/r2)^2 (10)&#160; &#160; &#160;<----- v=rw1,5=0.5 X w1, w1=10
w2=(0.5/0.3)^2(10)
w2=27.77777778 rad s-1
v2=r2w2
v2=0.3(27.77777778)
v2=8.3333 ms-1

vick5821

How bout c)??

cheesit92

work done= gain in kinetic energy,
(&#189I2w2^2 - (&#189I1w1^2  =(&#189mr2^2 w2^2  - (&#189mr1^2 w1^2
                                      =(&#189(0.3)(0.3)^2 (27.78)^2 -(&#189(0.3)(0.5)^2 (10)^2
                                      =6.6668J

vick5821

Gravitational Ques!!
1.A rocket is launched vertically from the surface of the Earth at speed 25 km s-1. Determine its speed when it escapes from the gravitational field of the Earth.

How ar??
URGENT!!!!

vick5821

A particle moving along x axis in SHM starts from the origin at t=0 and moves to the right. The amplitude of the its motion is 2.00cm and the frequency is 1.50 Hz.
Calculate the maximum acceleration and the earliest time (t>0) at which the particle has this acceleration

Winterblade

a=-(w^2)(x)
Amplitude peaked at 0.02m
Frequency= 1.50Hz
a=-(2piF)(0.02)
  =-1.78ms^-2

T=period for one complete cycle.
  =2pi/w
  =2pi/2piF
  =1/1.5
  =2/3s

Earliest peak occur at 1/4 T
1/4x2/3= 1/6s

I am not certain about that...Sorry if i make any mistakes

vick5821

yes..I think u r correct..the answer given is wrong..thx for helping!!

Winterblade

>.<...What is the answers?

nkrealman

关于 static...
ladder with centre of mass resting on wall，斜放着。
为什么頭尾的normal reaction 是斜上去？

Winterblade

回复 430# nkrealman


    I think is due to the rough surface of the floor and wall....

whyyie

回复 430# nkrealman

斜上去? 不是perpendicular吗?

Winterblade

I thought perpendicular normal reaction is for smooth surface?

Log

关于 static...
ladder with centre of mass resting on wall，斜放着。
为什么頭尾的normal reaction 是 ...
nkrealman 发表于 31-8-2010 03:14 PM

可以这样想for ladder resting on a rough surface
actually the reaction firstly is acting perpendicularlly, but there exists a frictional force,so by resolving, u will get an inclined reaction force
for smooth surface,the  reaction is normal coz no frictional force.

Winterblade

回复 434# Log

vick5821

the answer for the ques I postred is 1/2s, i think its wrong la..haha
hey, another question here

two waves in a long string are given by
y1 = 0.02m sin ( 40t - x/2) and y2 = 0.02m sin (40t + x/2)
where y1,y2 and x are in metres and t is in seconds
Calculate the maximum displacement of an element in the string at x=0.40m

I dun quite understand this ques..how to do??thanks for the help!!

Log

the answer for the ques I postred is 1/2s, i think its wrong la..haha
hey, another question here

...
vick5821 发表于 2-9-2010 12:51 PM

y1=0.02sin(40t-x/2)
y2=0.02sin(40t+x/2)
resultant wave in the string y=y1+y2
                                         =0.02sin(40t-x/2) + 0.02sin(40t+x/2)
                                         = 0.02 (2  sin 40t cos x/2)
                                        y = 0.04  cos(x/2) sin (40t)
    amplitude=0.04 cos x/2
x=0.40m , so, max diplacement= 0.04 cos (0.40/2)= 0.04 m

vick5821

I dun understand why do like that?? why just take the front part of the equation??

Log

就好像y=a sin bt ,    a=amplitude ,  (y and t are variables)
  etc.      y=2a/c sin (bt +d)               2a/c = amplitude
              y= 2 cos x sin bt             ,  2 cos x=amplitude
so, y= 0.04 cos (x/2 ) sin 40t
0.04 cos (x/2) = amplitude ,   x=distance of a particle measured from the origin = constant

Winterblade

回复 439# Log


    Uhmm how about  resultant wave eqn is y= cos 2(pi)t ((f1-f2)/2)  sin 2(pi)t ((f1+f2)/2)....What is the amplitude?