Problem #1
A particle undergoes rotational motion because it obtains a moment of force of 30 Nm. If the angular acceleration that occurs is 3 rad/s2, determine the moment of particle inertia.Answer:
Known:
Angle acceleration, α = 3 rad.s-2
Torque, τ = 30 Nm
Then the moment of particle inertion is
I = τ / α = 30/3 = 10 kg.m2
Problem #2
On a flywheel whose inertia moment of 4.0 kg.m2 is done a 50 mN moment. Six seconds after starting to rotate at an angular velocity of 40 rad/s the flywheel has gone round as far. . . ?
Answer:
Known:
Moment of inertia, I = 4.0 kgm2,
Torque, τ = 50 mN,
Time, t = 6 seconds
and angular velocity, ω = 40 rad/s.
The angular acceleration experienced by the flywheel is
α = τ/I = 50 mN/4.0 kgm2 = 12.5 rad/s2
The round taken by the crazy wheel for 6 seconds is
Δθ = ω0t + ½αt2 = 40 rad/s x 6 s + ½ (12.5 rad/s)(6 s)2 = 465 rad
Problem #3
A bicycle wheel with radius 0.3 m can rotate against a fixed axis. The moment of inertia of the wheel against that axis is 0.5 kgm2. A constant force of 2 N is worked tangentially on wheel rim for 0.6 s. The wheel starts to move from rest. After 0.6 s the angular velocity of the wheel is. . . ?
Answer:
Known:
Wheel radius, R = 0.3 m,
moment of inertia with respect to axis, I = 0.5 kgm2,
constant force F = 2 N,
time t = 0.6 s.
Then the acceleration of the wheel angle can be obtained from
∑τ = Iα
FR = Iα
(2 N) (0.3 m) = 0.5 kgm2α
α = 1.2 rad/s2
then the angular velocity of the wheel is
ω = ω0 + αt
ω = 0 + (1.2 rad/s2)(0.6 s) = 0.72 rad/s2
Problem #4
The object in the diagram below is on a fixed frictionless axle. It has a moment of inertia of I = 50 kgm2 . The forces acting on the object are F1 = 100 N, F2 = 200 N, and F3 = 250 N acting at different radius R1 = 60 cm, R2 = 42 cm, and R3 = 28 cm. Find the angular acceleration of the object.
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| Fig.1 |
Known:
moment of inertia, I = 50 kgm2,
The forces acting on the object are F1 = 100 N, F2 = 200 N, and F3 = 250 N acting at different radius R1 = 60 cm, R2 = 42 cm, and R3 = 28 cm.
Since the axle is fixed we only need to consider the torques and use Στz = Izαz . Each of the forces is tangential to the object, i.e R and F are at 90º to one another. Recall that clockwise torques are negative or into the paper in this case.
Στz = Izαz
R1F1 + R2F2 + R3F3 = Iα
(0.6 x 100) + (0.42 x 200) + (0.28 x 250) = 50α
α = 1.88 rad.s-2
Problem #5
A ball (mass m1 = 3 kg) and a block (mass m2 = 2 kg) are connected to a rope (negligible mass) through a pulley (radius R = 20 cm and moment of inertia Ipm = 0.5 kg.m2) as shown in the picture. Both objects move at v speed and the block moves without friction with the floor. The linear acceleration of the two objects is. . . .?
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| Fig.2 |
Known
Mass, m1 = 3 kg, m2 = 2 kg
moment of inertia, I = 0.5 kgm2,
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| Fig.3 |
m1g - T1 = m1a
3 kg x 10 m/s2 - T1 = 3a
30 - T1 = 3a (*)
We review the block (m2)
T2 = m2.a
T2 = 2a (**)
We get the equation (*) and (**)
T1 - T2 = –5a + 30 (***)
We review the pulley
∑τ = Iα
(T1 - T2)R = 0.5a/R
Then from (***), we get it
(–5a + 30) 0.2 = 0.5a/0.2
-a + 6 = 5a/2
a = 12/7 = 1.7 m/s²
Problem #6
A rod with a mass of 2M and length L is connected to the wall by a hinge, where the free stem rotates. A horizontal rope connected to the end of the rod holds the rod at an angle θ to the vertical direction, as shown in the figure. If the rope suddenly breaks, determine the acceleration angle of the rod!
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| Fig.4 |
Of course the stem rotates on the hinge (O axis), so the forces acting on the rod are shown in the figure below,
The torque at point O is caused by the gravity of the rod w because when the rope breaks the tension of the rope is zero (T = 0) and the contact force of the hinge and rod ie V and H does not cause a moment of force at O.
Style w, moment arm à OR = ½ L sin θ
Then, the resultant moment of force at point O is
ΣτO = Iα
the moment of stem inertia at one end is I = 1/3 ML2
w x OR = (1/3 ML2) α
(w x 1/2 L sin θ) = (1/3 ML2) α
(Mg x 1/2 L sin θ) = (1/3 ML2) α
α = (3g sin θ)/2L
Problem #7
A rope is wrapped around a solid cylindrical drum. The drum has a fixed frictionless axle. The mass of the drum is 125 kg and it has a radius of R = 50.0 cm. The other end of the rope is tied to a block, M = 10.0 kg. What is the angular acceleration of the drum? What is the linear acceleration of the block? What is the tension in the rope? Assume that the rope does not slip.
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| Fig.5 |
The forces acting directing on the block are weight and tension. Presumably the block will accelerate downwards. The only force directly acting on the drum which creates a torque is tension. Note that ropes, and therefore tensions, are always tangential to the object and thus normal to the radius. The other forces acting on the drum, the normalfrom the axle and the weight, both act through the CM and thus do not create torque. The drum accelerates counterclockwise as the block moves down.
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| Fig.6 |
Referring to the table of Moments of Inertia, we find that I = ½mR2 for a solid cylinder.
We review the block (M)
-T + Mg = Ma
T = Mg – Ma
We review pulley
T = Iα = (1/2 MR2)α
Our second is
RT = ½mR2(a/R),
or when we simplify
T = ½ma.
Putting this result into the first equation yields
a = Mg/[M + ½m] = 1.353 m/s2 .
Thus α = a/R = 2.706 rad/s2
As well,
T = ½ma = ½mMg/[M + ½m] = 84.56 N.
Problem #8
A winch has a moment of inertia of I = 10.0 kgm2 . Two masses M1 = 4.00 kg and M2 = 2.00 kg are attached to strings which are wrapped around different parts of the winch which have radius R1 = 40.0 cm and R2 = 25.0 cm. (a) How are the accelerations of the two masses and the pulley related? (b) Determine the angular acceleration of the masses. Recall that each object needs a separate free body diagram. (c) What are the tensions in the strings?
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| Fig.6 |
Answer:
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| Fig.7 |
(b) If we use the relationships from part (a), we can rewrite the equations in the table as
T1 = M1g – M1R1α, and
T2 = M2g + M2R2α.
We use these results to eliminate T1 and T2 from the torque equation
R1T1 – R2T2 = Iα
R1 [M1g – M1R1α] – R2 [M2g + M2R2α] = Iα
R1M1g – R2M2g = [I + M1(R1)2 + M2(R2)2]α.
Thus we find
α = g(R1M1 – R2M2 )/[I + M1(R1)2 + M2(R2)2] = 1.002 rad/s2 .
(c) Using this results, and our previous equations for the tension in each string, we find
T1 = M1g – M1R1α = 38.60 N, and
T2 = M2g + M2R2α = 20.12 N
Problem #9
A rope connecting two blocks is strung over two real pulleys as shown in the diagram below. Determine the acceleration of the blocks and angular acceleration of the two pulleys. Block A is has mass of 10.0 kg. Block B has a mass of 6.00 kg. Pulley 1 is a solid disk, has a mass of 0.55 kg, and a radius of 0.12 m. Pulley 2 is a ring, has mass 0.28 kg, and a radius of 0.08 m. The rope does not slip.
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| Fig.8 |
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| Fig.9 |
Translational motion block A:
mAg – T1 = mAa
Translational motion block B:
T2 – mBg = mBa
Rotational motion pulley 1:
Στ1 = Idiskα1
(T1 – T2)R1 = Idiskα1
Rotational motion pulley 2:
Στ2 = Ihoopα2
(T2 – T3)R2 = Ihoopα2
In addition to the equations we have found above, we also know that the tangential acceleration of the pulleys is the same as the acceleration of the rope. Thus the angular acceleration of each pulley is related to a by α1 = a/R1 and α2 = a/R2 . Examining a table of Moments of Inertia reveals that Idisk = ½Mdisk (R1)2 and Idisk = Mhoop (R2 )2 . Using this information allows us to rewrite the equations as
T1 = MAg MAa (1),
T1 – T2 = ½Mdiska (2),
T2 – T3 = Mhoopa (3), and
T3 = MBg + MBa (4).
If we add equations (2) and (3) together, we get
T1 – T3 = (½Mdisk + Mhoop)a .
Then equations (1) and (4) can be used to eliminate T1 and T3 from the above
[MAg – MAa] – [MBg + MBa] = (½Mdisk + Mhoop)a .
Collecting terms involving a and rearranging yields,
a = (MA – MB)g/(MA + MB + ½Mdisk + Mhoop) = 2.370 m/s2 .
Using the above result, we find the angular accelerations
α1 = a/R1 = 19.75 rad/s2
and α2 = a/R2 = 29.63 rad/s2
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