Transcript Random Problems

Random Problems
The following is a random collection of
problems. There is no special significance of
the appearance or the order of these problems.
A grinding wheel accelerates from rest to an
angular velocity of 675 rev/min in 1.25 min.
Determine the:
(a) angular acceleration in rad/s2
ωi = 0
ωf = 675 rev/min
ωf = ωf + α Δt
Δt = 1.25 min
.
rev × min × 2π rad /
α = 675 min
60 s
1 rev
60 s
(1.25 min × min ) = 0.942 rad/s2
(b) angular displacement after 1.25 min
Δθ = ωiΔt + ½αΔt2
rad
60 s 2
Δθ = 0 + ½ × 0.942
× (1.25 min × min )
2
s
Δθ = 2650 rad
(c) angular displacement during the second
30.0 s interval.
Δθ = ωiΔt + ½αΔt2
Δθ60 – Δθ30 = ½ × 0.942 rad/s2 × (60.0 s)2 ½ × 0.942 rad/s2 × (30.0 s)2
Δθ60 – Δθ30 = 1270 rad
A 1 kg mass is hung from a rope which is
fastened to the ceiling.
What downward force acts on the 1 kg mass?
The weight of the 1 kg mass.
According to Newton’s 3rd Law, what is the
reaction force?
The reaction force is the 1 kg mass pulling
straight up on the earth.
Why do these forces not cancel out?
An action-reaction pair of forces are equal in
magnitude, opposite in direction, but act on
different objects.
A grinding wheel has a constant angular velocity
of 55 rev/min.
(a) Determine the angular velocity in rad/s.
ω = 55 rev × 2π rad × 1 min
rev
min
60 s
= 5.76 rad/s
(b) Determine the angular acceleration in rad/s2.
If the angular velocity is constant, then the
angular acceleration is 0.
A car with an initial velocity of 30.0 m/s
accelerates for 8.0 s at the rate of 7.0 m/s2.
(a) What is the car’s total displacement?
vi = 30.0 m/s
a = 7.0 m/s2
Δt = 8.0 s
Δx = viΔt + 1/2aΔt2
Δx = 30.0 m/s × 8.0 s + ½ × 7.0 m/s2 × (8.0 s)2
Δx = 464 m
(a) What is the car’s displacement during the
eighth second?
Δx = viΔt + 1/2aΔt2
Δx = 30.0 m/s × 7.0 s + ½ × 7.0 m/s2 × (7.0 s)2
Δx = 382 m
Δx = 464 m - 382 m = 82 m
A 2.0 kg mass is suspended from a spring.
When the mass is set in motion and air
resistance is negligible.
Will the total energy of the system be
conserved?
Yes, mechanical energy is always conserved in
the presence of conservative (non-dissipative)
forces such as gravity and the restoring force
within a spring.
How many forms of potential energy are
present in this scenario?
There are two forms, elastic potential (½kx2)
and gravitational (mgΔh) potential energy.
A net force of 15.0 N acts on an object with a
mass of 2.5 kg.
(a) If the object starts from rest, how much
work is done in the first three seconds?
Fnet = 15.0 N
W = F × x × cosθ
Fnet = ma
m = 2.5 kg
vi = 0
Fnet
a= m
.
15.0 N
=
= 6.0 m/s2
2.5 kg
Δx = x0 + viΔt + ½aΔt2 = 0 + ½ × 6.0 m/s2 × (3.0 s)2
Δx = 27 m
W = F × x × cosθ
W = 15.0 N × 27 m = 4.0 x 102 J
(b) How much work is done during the seventh
second?
Δx7 = x0 + viΔt + ½aΔt2 = 0 + ½ × 6.0 m/s2 × (7.0 s)2
Δx7 = 150 m
Δx6 = x0 + viΔt + ½aΔt2 = 0 + ½ × 6.0 m/s2 × (6.0 s)2
Δx6 = 110 m
W = 15.0 N × 150 m – 15.0 N × 110 m = 6.0 x 102 J
A ferrous wheel rotates at 0.25 rad/s.
(a) Determine the number of revolutions in 2.5
min.
ω = 0.25 rad/s
Δt = 2.5 min
ω = Δθ/Δt
Δθ = 0.25 rad/s × 2.5 min × 60 s/min ×
1 rev/2π rad = 6.0 rev
(b) Determine the time needed to stop the
ferrous wheel if it decelerates at
-0.010 rad/s2.
α = -0.010 rad/s2
ωf = ωi + αΔt
Δt = 0 – 0.25 rad/s/-0.010 rad/s2 = 25 s
A soccer ball with a mass of 5.00 kg rolls at
9.00 m/s. If the soccer ball does not slide along
the ground, determine its:
(a) linear kinetic energy
m = 5.00 kg
v = 9.00 m/s
KE = ½mv2 = ½ × 5.00 kg × (9.00 m/s)2
KE = 202 J
(b) rotational kinetic energy given that
I = 2/5mr2.
KE = ½Iω2 = ½ × 2/5 × m × r2 × v2/r2
KE = mv2/5 = 5.00 kg × (9.00 m/s)2/5 = 81.0 J
(c) Was it important to be told that the soccer
ball did not slip or slide along the way?
Yes, because if it had, you would to be told for
long the ball slid to compute the KErot.
(d) What is the significance of not being given
the radius of the soccer ball?
The radius is not important because it cancels
out in the calculation.
This fact is important to keep in mind.
Sometimes it is helpful to simplify the
calculation before performing the calculation.
In this case, you would not have been able to
do the problem because of the missing radius.
Pete is driving his car eastward at the same
time, his brother Repeat, is driving westward.
Because they were so excited to see each
other, they forgot to break and had a head on
collision. Pete and his car had a mass of m and
Repeat and his car had a mass of 5m.
(a) If Repeat’s original velocity is v, what was
Pete’s velocity before the collision?
mP = mP
vP = ?
mR = 5mP
vR = v
Δp = 0
pi = p f
mPvP + mRvR = mPvP’ + mRvR’
mPvP + 5mPvR = 0
vP = -5vR
(b) If during the force of impact, Pete exerted a
force of F on Repeat, what force did Repeat
exert on Pete?
According to Newton’s 3rd Law, for every force
there is an equal but opposite force, therefore,
Repeat exerted on force, F, on Pete.
A 6.50 kg mass is dropped from a height of
0.575 m striking a vertically placed spring. If
the elastic constant of the spring is
2.50 x 103 N/m, how much did the spring
compress?
m = 6.50 kg
h = 0.750 m
k = 2.50 x 103 N/m
ΔE = 0
(PEe + PEg)T = (PEe + PEg)B
(0 + mgΔh)T = (½kΔx2 + 0)B
x = (2 × 6.50 kg × 9.80 m/s2 × 0.575 m/
(2.50 x 103 N/m))½
x = 0.171m
A baseball with a mass of 0.145 kg starts rolling
from the top of an incline which is 3.0 m high. If
the incline is 12.0 m long, determine the:
(a) linear velocity of the baseball when it
reaches the bottom.
h = 3.0 m
l = 12.0 m
m = 0.145 kg
ΔE = 0
(PEg + KEt + KEr )T = (PEg + KEt + KEr )B
mgΔh + 0 = ½mv2 + ½Iω2
mgΔh = ½mv2 + ½ × 2/5 × m × v2/r2
9.80 m/s2 × 3.0 m = 7/10 × v2
v = 6.5 m/s
(b) the angular kinetic energy when it
reaches the bottom.
ΔE = 0
(PEg + KEt + KEr )T = (PEg + KEt + KEr )B
mgΔh + 0 = ½mv2 + ½Iω2
½Iω2 = 0.145 kg × 9.80 m/s2 × 3.0 m –
½ × 0.145 kg × (6.5 m/s)2
½Iω2 = 1.2 J