66
kaymant
·2009-03-30 12:22:00
(a) Let the plank move by distance x, then the center of mass of cylinder moves by x/2.
Hence, by work energy principle
Fx=\dfrac{1}{2}m_2v_2^2+\dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}\cdot \dfrac{1}{2}m_1R^2\omega^2
Here, v1 and v2 are, respectively, the speeds of c.m of cylinder and the plank, and ω is the angular speed of the cylinder. Since
v_2=2v_1, \qquad \textrm{and}\quad \omega = \dfrac{v_1}{R}=\dfrac{v_2}{2R}
we get
Fx=\dfrac{1}{2}m_2v_2^2 + \dfrac{1}{2}m_1\dfrac{v_2^2}{4}+ \dfrac{1}{2}m_1\dfrac{v_2^2}{8}
i.e.
Fx=\dfrac{1}{2}v_2^2\left(m_2+\dfrac{3}{8}m_1\right)
Hence,
v_2=\sqrt{\dfrac{2Fx}{m_2+\frac{3}{8}m_1}}
Comparison with v=√2as gives the acceleration a2 of the plank as
a_2=\dfrac{F}{m_2+\frac{3}{8}m_1}=\dfrac{8F}{8m_2+3m_1}
Hence, the acceleration of the center of mass of the cylinder
a_1=\dfrac{a_2}{2}=\dfrac{4F}{8m_2+3m_1}
(b) Let f_2 be the force of friction between the plank and the cylinder, then
F-f_2=m_2a_2, which give
f_2=F-m_2a_2=F-\dfrac{8m_2F}{8m_2+3m_1}=\dfrac{3m_1F}{8m_2+3m_1}
Let f1 be the friction force between the cylinder and the ground, then f1 must act forward. Hence, we get
f1+f2 = m1a1
from where we get
f_1=m_1a_1-f_2=\dfrac{4m_1F}{8m_2+3m_1}-\dfrac{3m_1F}{8m_2+3m_1}=\dfrac{m_1F}{8m_2+3m_1}
3
iitimcomin
·2009-03-30 07:36:52
F - f = m2a2 ...(1)
(f-f2)R = Iα .........(2)
f + f2 = m1a1 ........(3)
a1 = Rα .........................(4)
a1 + a2 + Rα= 0 ...................(5)
PLS CHECK IF EQXNS ARE RITE SIR
ive taken alpha in anti clock dir and a1 and a2 rightward............
11
virang1 Jhaveri
·2009-03-30 07:46:40
Magnitude and direction of friction
Friction at the top of the ball
The direction is opposite to the force applied.
Force of friction = Ï
1m2g providing anti-clockwise motion
Friction at the bottom
The direction is to provide clockwise rotation.
Force of friction =Ï
2(m1+m2)g
3
iitimcomin
·2009-03-30 07:48:02
no virag .......dont think ur right abt the mag.