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= mssg - xmcg = 0so that
x = = = L.Thus, the child must sit L/4 away from the fulcrum.
= mssg - mcg = 0which gives
mc = mss = (30) = 15 kg.
A 10 m ladder weighs 50 N and balances against a smooth wall at 60 o to the horizontal. If the ladder is just on the verge of slipping, what is the static coefficient of friction between the floor and the ladder?
The force diagram for the ladder is shown in Figure 8.4.
We need to impose both equilibrium conditions: i) = 0 and ii) = 0.
|Fx = 0||=||fs - N1|
|0||=||N2 - N1|
|[2mm]Fy = 0||=||N2 - mg|
0 = mg - N1.
|= 0||=||- mg(L/2)cos + N1Lsin|
|=||= = 0.29|
Calculate the x and y positions of the center of mass of the following three objects where m1 = 1.0 kg , m2 = 2.5 kg , and m3 = 4.0 kg . See Figure 8.5.
|=||= 1.6 m|
|=||= 0.27 m.|
Calculate the moments of inertia of the following 4 masses where m1 = 2.0 kg and m2 = 1.0 kg. See Figure 8.6.
a) About the axis through O (center) perpendicular to the page.
b) About an axis joining two of the masses.
Note: The moment of inertia depends strongly on the axis chosen.
|=||2m1(22) + 2m2(32)|
|=||8m1 + 18m2|
|=||34 kg m 2.|
First consider the x -axis (i.e. the one that passes through the
|=||m1(22) + m2(02) + m1(22) + m2(02)|
|=||2m1(22) = 16 kg m 2.|
For the y -axis (i.e. the one that passes through the two
|=||m1(0) + m2(32) + m1(0) + m2(32)|
|=||2m2(32) = 18 kg m 2.|
A round object, mass m , radius r and moment of inertia IO , rolls down a ramp without slipping as shown in Fig. 8.6 below. a) If the ramp is at an angle to the horizontal, find an expression for the acceleration of the center of mass of the object in terms of m,r,I0 and . b) What is the velocity of the center of mass of a solid sphere, IO = mr 2, if it starts from rest and rolls down a 4 m ramp inclined at 30 o to the horizontal?
|Fx = macm||=||mgsin - f|
|[2mm]Fy = 0||=||N - mgcos .|
For the rotational motion we have:
|= IO||=||- fsr|
acm = - at = - r.Thus we can write
fs = .Subsituting this into the above equations gives:
acm = gsin - IOwhich can easily be solved to yield:
Note that the mass and radius of the object cancel out of the final expression.
acm = gsinas expected.
Do the above problem (part b) using energy conservation.
|Wnc = 0||=||KEt + KEr + PEg|
|=||(mvcm2 - 0) + (I - 0) + (0 - mgh)|
|0||=||mvcm2 + I - 2mgh.|
h = Lsin (fromthediagram)Also we have that:
|I sphere||=||mr 2.|
|=||mvcm2 + (mr 2) - 2mgLsin|
|=||5vcm2 + 2vcm2 - 10gLsin|
|vcm||=||= 5.3 m/s|
A cylinder with moment of inertia I = 34 kg m 2 rotates at = 1.4 rads/s . It drops onto a second identical cylinder, and they stick together end to end. a) Calculate the final angular velocity of the two cylinders. b) Calculate the kinetic energy lost in the process.
|I + I||=||I + I|
|=||= 0.7 rads/s.|
|KEr||=||KErf - KEri|
|=||(I + I) - I|
|=||I - I|
|=||- 16.7 J|
A student stands on the center of a rotating platform that has frictionless bearings. He has a 2.0 kg object in each hand, held 1.0 m from the axis of rotation of the system. The system is initially rotating at 10 rpm. Determine a) the initial angular velocity in radians per second, b) the angular velocity of the sytem in radians per second after the objects are brought to a distance of 0.20 m from the axis of rotation, c) the change in the rotational kinetic energy of the system as the objects are pulled closer to the center of rotation. d) What causes the increase in rotational kinetic energy? (Assume that the moment of inertia of the platform + student remains constant at 1.0 kg m 2.)
|10 rev/min||=||(10 rev/min )(2 rads/rev )(1 min /60 s )|
|=||1.05 rad /s.|
|Ii||=||I (student + platform) + mri2 + mri2|
|=||1.0 kg m 2 + (2.0 kg )(1.0 m )2 + (2.0 kg )(1.0 m )2|
|=||5.0 kg m 2|
|If||=||I (student + platform) + mrf2 + mrf2|
|=||1.0 kg m 2 + (2.0 kg )(0.2 m )2 + (2.0 kg )(0.2 m )2|
|=||1.16 kg m 2.|
|Ii = If.|
|=||x 1.05 rads/s|
|KE||=||KEf - KEi|
|=||If - Ii|
|=||(1.16 kg m 2)(4.5 rads/s )2 - (5.0 kg m 2)(1.05 rad /s )2|
|=||11.8 J - 2.75 J|