Problem #1
A solid cylinder with a mass of 10 kg is above a rough surface drawn by force F = 50 N as shown in the following figure!
Determine the acceleration of the cylinder motion if the fingers are R = 40 cm!
Answer:
Review the forces on the cylinder:
∑τ = Iα
(F – f)R = ½ mR2(a/R)
F – f = ½ ma (1)
translations motion:
∑F = ma
F + f = ma (2)
Add the equations (1) and (2), we get
2F = (3/2)ma
F = ¾ ma
50 = ¾ (10)a
a = 6.67 m.s-2
Problem #2
A yoyo has a mass M = 4.0 kg, a moment of inertia I, and an inner radius r = 2.0 cm. A string is wrapped around the inner cylinder of the yoyo. A person ties the string to his finger and releases the yoyo. As the yoyo falls, it does not slip on the string (i.e. the yoyo rolls). Find the acceleration of the yoyo if Iyoyo = 0.028 kg.m2.
Answer:
The yoyo is said to roll without slipping. That phrase means that the angular acceleration of the yoyo about its CM is related to its linear acceleration by a = Rα. Note that since the string is tied around the inner cylinder; it is that radius which figures into the relation.
Since αcm = a/r, we have two simple equations
For translations motion
∑F = ma
mg – T = ma, or
T = mg – ma,
For rotational motion
∑τ = Iα
Tr = Iα = Ia/r, or
T = Ia/r2.
Eliminating T from the first equation yields,
a = mg/[m + I/r2]
with the value that we know we have
a = 4.0 x 10/(4.0 + 0.028/0.022) = 0.54 m.s2
Problem #3
A solid cylinder rolls down an inclined plane without slipping. The incline makes an angle of 25.0 to the horizontal, the coefficient of static friction is µs = 0.40, and Icyl = ½MR2. Hint you may not assume that static friction is at its maximum!
(a) Find its acceleration.
(b) Find the angle at which static friction is at its maximum, at just above this angle the object will start to slip.
Answer:
The cylinder is said to roll without slipping. That phrase means that the angular acceleration of the ball about its CM is related to its linear acceleration by a = Rα.
The type of friction is static since we are told that the cylinder is rolling without slipping. The only point left to resolve is in which direction it points. Since friction creates the only torque, and we have decided that α is forward, then friction must be up the incline. Only point to be careful about is that in rolling problems, one seldom is dealing with the fsMAX unless it is explicitly stated.
∑Fx = mg sin θ – fs = ma (*)
review cylinder translation motion on the y axis:
∑Fy = N – mg cos θ = 0 (*)
review cylinder rotation motion:
∑τ = Iα
fsR = ½ mR2(a/R)
fs = ½ ma (***)
from the equation (*) and (***) we get:
mg sin θ – ½ ma = ma
a = 2g sin θ/3
a = 2/3 x 9.8 x sin 250 = 2.764 m/s2
(b) We can use this result with fs = ½ma, to get an expression for fs,
fs = (1/3)mg sinθ (1)
However, the maximum value of fs is μsN. The second equation gives N = mg cosθ, so
fs = μsmg cosθ (2)
Using (1) and (2) to eliminate fs, yields
(1/3)mg sinθ = μsmg cosθ .
Using the identity tanθ = sinθ/cosθ, we get
θ = tan–1(3μs) = 50.20 .
This is the angle at which the cylinder would start to slip as it moved down the incline.
Problem #4
A person pulls a heavy lawn roller by the handle with force F = 24.0 N so that it rolls without slipping. The handle is attached to the axle of the solid cylindrical roller.
The handle makes an angle θ = 37.10 to the horizontal. The roller has a mass of M = 3.0 kg and a radius R = 10 cm. The coefficients of friction between the roller and the ground are μs = 0.5 and μk = 0.2.
(a) Find the acceleration of the roller.
(b) Find the frictional force acting on the roller.
(c) If the person pulls too hard, the roller will slip. Find the value of F at which this occurs
Answer:
First we draw the FBD. Clearly we have F, weight, and a normal force acting on the roller. As well, there must be static friction since we have rolling without slipping. It is not at a maximum since the roller only starts to slip in part (c). The direction of fs must be to the right, since the force F pulls the roller into the ground, the ground pushes back. We assume the accelerations are as shown, left and ccw. Note that these are consistent.
review cylinder translation motion on the x axis:
Σ Fx = max
− Fcos θ + fs = − Ma
review cylinder translation motion on the x axis:
ΣFy = may
N + Fsin θ − Mg = 0
review cylinder rotation motion:
Στcm = Icmαcm
Rfs = Iα
We also know that a = Rα and I = ½MR2. We substitute these relationships into the torque equation
Rfs = ½MR2(a/R)
This yields an equation for fs,
fs = ½Ma
We take this result and substitute it into the xcomponent
Equation
− Fcos θ + ½Ma = − Ma (3)
Solving for a, as required in part (a), yields
a = (2/3)(F/M) cos θ (4)
a = (2/3)(24.0/3.0) cos 37.10 = 4.27 m.s−2
(b) We can also find fs, as required in part (b), by substituting (4) back into Eqn. (2);
fs = (1/3) Fcos θ (5)
fs = (1/3) x 24.0 x cos 37.10 = 6.40 N
(c) asks us when the roller will slip. This occurs when fs = fsmax. Now we know fs
max = μsN. We find N from the ycomponent equation, so
fsmax = μs[Mg – Fsin θ] (6)
Equating Eqns. (5) and (6) will tell us the value of F at which slipping occurs
(1/3)Fcos θ = μs [Mg – Fsin θ] (7)
We get F by itself on the lefthand side
F[(1/3)cos θ + μs sin θ] = μsMg (8)
So
F = 3μsMg/[cos θ + 3μs sin θ]
F = 3 x 0.5 x 3.0 x 9.8/[cos 37.10 + 3 x 0.5 sin 37.10] = 8.65 N
Problem #5
A yoyo of Mass M, moment of inertia I, and inner and outer radii r and R, is gently pulled by a string with tension T as shown in the diagram below.
(a) Find the acceleration.
(b) Find the friction acting on the yoyo.
(c) At what value of T will the yoyo begin to slip?
Answer:
We know there is a normal and weight acting on the yoyo but these do not create torques as they operate on or through the Centre of Mass. There is nonmaximum static friction acting but we must determine it's direction. First if there were no T, there would be no friction. T is twisting the yoyo counterclockwise pushing the yoyo into the surface. The surface reacts by pushing back. The FBD looks like. Here I have guessed that the yoyo will roll backwards. Notice that my choice of a and α are consistent.
review yoyo translation motion on the x axis:
Σ Fx = max
T − fs = − Ma
review cylinder translation motion on the x axis:
ΣFy = may
N − Mg = 0
review cylinder rotation motion:
Στcm = Icmαcm
rT − Rfs = Iα
We also know that a = Rα.
The torque equation becomes can be solved for fs
fs = (r/R)T − (I/R2)a (1)
We substitute this into the x−component equation
T − [(r/R)T − (I/R2)a] = − Ma (2)
We bring the term involving a from the left to the right and solve for a in terms of T,
T(R−r)/R = − (M + I/R2)a (3)
Or
a = −T(R−r)R/[MR2 + I] (4)
The fact that a is negative tells me that my guess about the direction of a and α are wrong. The acceleration is forward and counterclockwise.
Substituting Eqn. (4) into Eqn. (1), we find
fs = [MRr + I]T/[MR2 + I]
Since fs is positive, it must be have been chosen in the right direction.
Now as we see from Eqn. (5), as T increases so does fs. We know fs £ μsN or fs £ μsMg in our case
when we make use of the y−component equation. Thus we have a limit on T
[MRr + I]T/[MR2 + I] ≤ μsMg
This yields the result
T ≤ μsMg[MR2 + I]/[MRr + I]
If T is any bigger, the yoyo will slip.
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