Problem#1
A solid cylinder is homogeneous with a radius of 20 cm with a mass of 2 kg which is at the top of a rough inclined plane at an altitude of 1.5 m. if the cylinder slides down the inclined plane with a slope of 30 °. determine the speed of the cylinder when it arrives at the base of the inclined plane!Answer:
By using the conservation law of energy, we can solve this problem.
Mechanical energy at the top of the inclined plane = mechanical energy at the base of the inclined plane
EMtop = EMbase
EPtop + (EKrot + EKtrans)top = EPbase + (EKrot + EKtrans)base
Mgh + 0 = 0 + ½Iω2 + ½Mv2,
Karena I = ½ MR2 dan ω = v/R
Mgh = ½ (½MR2)(v/R)2 + ½Mv2
gh = ¾v2
10 m/s2 x 1,5 m = ¾v2
v = 2√5 m/s
Problem#2
A thin hollow ball with a mass m and a finger r roll with linear velocity v along a horizontal plane without slipping. If the moment of spherical inertia is 2mR2/3, the kinetic energy of the ball is. . . .?
Answer:
The ball rolls then the ball's kinetic energy is
EKTOTAL = EKTRANS + EKROT
EKTOTAL = ½ mv2 + ½Iω2
EKTOTAL = ½mv2 + ½(2mR2/3)(v/R)2 = mv2/3
Problem#3
A solid ball with a finger R and mass m is driven with an initial velocity of 3 m/s above an inclined plane which has a slope of 30 ° with a kinetic friction coefficient of 0.2 and a static coefficient of 0.3. When climbing the field, the ball always rolls and never slips. The maximum height reached by the ball is around. . . .?
Answer:
Mechanical energy at the top of the inclined plane = mechanical energy at the base of the inclined plane
EMtop = EMbase
EPtop + (EKrot + EKtrans)top = EPbase + (EKrot + EKtrans)base
Mgh + 0 = 0 + ½Iω2 + ½Mv2,
Karena I = 2MR2/5 dan ω = v/R
Mgh = ½ (2MR2/5)(v/R)2 + ½Mv2
gh = 7v2/10
9,8 m/s2 x h = 7(3 m/s)2/10
h = 0,64 m
Problem#4
A solid ball with a mass of 0.036 kg and a radius of 1.2 cm rolls down an inclined plane. the solid ball first moves at a speed of 0.50 m/s. The speed of the ball when the height is reduced by 14 cm is. . . . (gravitational acceleration g = 10 m/s2)
Answer:
the ball moves without slip, then mechanical energy is applied at the top of the inclined plane = mechanical energy at the base of the inclined plane,
EMtop = EMbase
EPtop + (EKrot + EKtrans)top = EPbase + (EKrot + EKtrans)base
Mgh + ½Iω02 + ½Mv02 = 0 + ½Iω2 + ½Mv2,
Karena I = 2MR2/5 dan ω = v/R
Mgh + ½ (2MR2/5)(v0/R)2 + ½Mv02 = ½ (2MR2/5)(v/R)2 + ½Mv2
gh + 7v02/10 = 7v2/10
10 m/s2 x 0,14 m + 7(0,5 m/s)2/10 = 7v2/10
v2 = 63/28
v = 1,5 m/s
Problem#5
A homogeneous stem with mass m and length L is given one end of the hinge so that the stem is freely rotating. The stem is raised so that its position is horizontal, then it is stretched from rest so that the rod rotates against an axis in the vertical plane. Angular velocity when the rod is in the vertical position is. . . .?
The central movement of rod mass applies the conservation law of mechanical energy, then
Mgh = ½Iω2
Mg(L/2) = ½ (ML2/3)ω2
ω = (3g/L)1/2
Problem#6
A ball with mass M and radius R has a moment of inertia I = 2MR2/ 5. The ball is freed from rest and rolls down the inclined plane without losing energy due to friction. The ball is thrown vertically up, out of the inclined plane as shown in the diagram reaching the maximum height above the point where the ball leaves the field.
Answer:
We solve this problem easily using the law of conservation of mechanical energy,
EMtop = EMbase
EPtop + (EKrot + EKtrans)top = EPbase + (EKroti + EKtrans)base
Mgh + 0 = 0 + ½ Iω2 + ½ Mv2
Baceuse I = 2MR2/5 dan ω = v/R
Mgh = ½ (2MR2/5)(v/R)2 + ½Mv2
gh = 7v2/10
vbase = (10gh/7)1/2
the ball will still move vertically upwards because it has kinetic energy and reaches max height
½ Mv2base = Mgymax
½(10gh/7) = gymax
ymax = 5h/7
Problem#7
A cylinder of mass M and radius R, on an incline of angle θ, is attached to a spring of constant K. The spring is not stretched. Find the speed of the cylinder when it has rolled a distance L down the incline.
Answer:
The only external force on the system, excluding gravity which is taken into account through gravitational potential energy, is static friction. For objects which roll across a stationary surface, static friction does no work, so here 0 = ΔE, where E is the observable or mechanical energy.
As the object rolls distance L down the incline, it loses gravitational potential energy while it gains both linear and rotational kinetic energy. Moreover, the spring gains Spring Potential Energy. Thus we have
0 = mgH + ½mv2 + ½Iω2 + ½KL2
The relationship between H and the L is H = Lsinθ. Since the object rolls without slipping, the angular velocity ω is related to the linear velocity v by ω = v/R. Furthermore, consulting a table of values, the moment of inertia of a cylinder about the perpendicular axis through the CM, is ½MR2. Our equation becomes
0 = mgLsinθ + ½mv2 + ½(½MR2)(v/R)2 + ½KL2
Isolating v2 terms, we find v2[½m + ¼m] = mgL sinθ ─ ½KL2. Solving for v we get
A small solid sphere of radius r = 1.00 cm and mass m = 0.100 kg at point A is pressed against a spring and is released from rest with the spring compressed 20.0 cm from its natural length. The spring has a force constant k = 20.0 N/m. The sphere rolls without slipping along a horizontal surface to point B where it smoothly continues onto a circular track of radius R = 2.00 m. The ball finally leaves the surface of the track at point C. Find the angle θ where the ball leaves the track. Assume that friction does no work. Hint find an expression for the speed of the sphere at point C. The moment of inertia of a sphere is I = 2/5mr2.
Answer:
The problem involves a change in height and speed and has a spring, so we apply the generalized Work─Energy Theorem., WNC = ΔE.
We are told to assume WNC = 0. The spring is compressed initially, so it loses spring potential energy. The ball increases both linear and rotational kinetic energy. The ball loses gravitational potential energy. Our equation is thus
0 = ─½kx2 + ½mv2 + ½Iω2 ─ mgh
We are told the moment of inertia, I, and we are told that the ball does not slip so ω = v/r. Using some trigonometry, the distance dropped is h = (R+r) ─ (R+r)cosθ. Using these results with equation (1) yields
0 = ─½kx2 + (7/10)mv2 ─ mg(R+r)[1 cosθ] (2)
We have two unknowns, v and θ. To proceed further we note that the ball loses contact with the surface. Recall that losing contact implies that N = 0. The normal is a force and to find forces we draw a FBD and apply Newton's Second Law. Since the ball moves in a circle, we are dealing with centripetal acceleration.
N ─ mgcosθ = mv2/(R+r)
Since N = 0, and after some rearranging, the force equation yields
v2 = (R+r)gcosθ
Substituting this into equation(2), we get
0 = ─½kx2 + (7/10)mg(R+r)cosθ ─ mg(R+r)[1 ─ cosθ]
Collecting terms with cos on the lefthand side yields
(17/10)mg(R+r)cosθ = ½kx2 + mg(R+r)
Solving for θ,
θ = cos─1{[½kx2 + mg(R+r)]/(17/10)mg(R+r)}
θ = 44.9º
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