A string is rolled around a cylinder ( m = 3.9 kg) as shown in the figure below.
ID: 2258757 • Letter: A
Question
A string is rolled around a cylinder (m = 3.9 kg) as shown in the figure below. A person pulls on the string, causing the cylinder to roll without slipping along the table. If there is a tension of 34 N in the string, what is the acceleration of the cylinder?
A board of length 1.7 m is attached to a floor with a hinge at one end as shown in the figure below. The board is initially at rest and makes an angle of 55
Explanation / Answer
ma=T-mg
a= (T-mg)/ m
a= 34-38.22/ 3.9
a= -4.22/3.9
a=-1.082 m/s^2
# Let the board length be L. Then the force of gravity acts at L/2, and the magnitude
of the torque on the board at the moment of release is the magnitude of the
cross-product
lr x Fl = (L/2)*(mg*sin(45)) = (sqrt(2) /4)*mgL.
Now torque and angular acceleration are related by torque = I*a, where I is the
mass moment inertia and a is the angular acceleration.
Now I believe that the moment of inertia for the board in this case is m*L^2 /3.
This would give us
a = torque / I = (sqrt(2) /4)*mgL / (m*L^2 /3) = (3/4)*sqrt(2)*(g/L),
which for g = 9.8 m/s^2 and L = 1.7 m gives a value of a = 6.1144 radians/s^2
# The net force which causes acceleration on m1 and m2 and an angualr accelration ? of the pulley is m2g - ?m1g = g {m2-? m1}
The angualr acceleration ? and linear acceleration a are related by a = r ?
m1a is the force causing an acceleraion on m1
m2a is the force causing an acceleration of m2 .
We have to find the force causing the pulley ro ratate with an angular acceleraion of ?
or in othere words to have a liear acceleration of a of a point on the rim of the pulley.
The torque F*r = I ? = I*a/r.
Hence F = I*a/r^2.
Equating the forces
g {m2-? m1} = a [ m1 +m2 + I/r^2}
Hence a = g {m2-? m1} / [m1 +m2 + I/r^2}
a. = g {m2-? m1} / [m1 +m2 + I/r^2}
The I of the pulley = m3r^2/2 and hence I/r^2 = m3/2
Hence a = g {m2-? m1} / [m1 +m2 + m3/2}
a = 9.8{21-0.14*12} / [12 +21 + 23/2} = 4.25 m/s^2
# for a disk, the moment of inertia is 1/2MR^2 where M is the mass of the disk and R is the radius
if we call R the radius of the entire CD and r the radius of the hole, we have that the moment of inertia of the CD is
1/2MR^2-1/2Mr^2= 1/2M(R^2-r^2)
if you know the value of M, you can calculate the actual value of the moment of inertia
if there were no hole, the moment would be 1/2MR^2, so the effect of the hole onthe moment of inertia ofthe CD can be estimated:
[1/2M(R^2-r^2)/1/2MR^2]x100% =
(R^2-r^2)/R^2x100% = [1-(r/R)^2]x100%
for r=0.7cm and R=3cm, we have
[1-(0.7/3)^2]x100% = 94.55% meaning the actual value of the moment of inertia is 94.55% of what it would be if there were no hole, so the hole contributes a loss of 5.45%
# the system is in static equilibrium so the sum of the torque and sum of the forces is zero.
the pivot point is the corner of the edge of the cube on the ground farthest from the person (because when it starts to tip, the cube will pivot about that point) so:
sum of torque = 0
gravity acts at center of mass (half of the edge length), torque applied by the person will be with lever arm .50 m and it will be negative because it is pushing the cube in a clockwise direction, Normal force is at the pivot point so its torque is zero, force of friction is also at the pivot point so its torque is zero (lever arm=0)
So: mg(.55/2) - F(.55)= 0
*note you don't even need the edge length because the force will equal mg/2
(35)(9.8)/2=F
F=171.5 N
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