Download Jumping problems

Jumping problems:
Two phases:
Jump with forces from gravity and leg muscles, acceleration upward.
Flight, animal has left ground, gravity only force, acceleration downward.
Treat these separately, each with it’s own acceleration.
Final velocity from jump is the initial velocity for the flight.
P2.53. When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of
0.50 mm.
a) What is the flea’s acceleration during the jump phase?
b) How long does the acceleration phase last?
c) If the flea jumps straight up, how high will it go?
Ignore air resistance here, which in reality would be very large and the flea will not reach this height.
Prepare: The motion has two parts: The jump, with an acceleration upward, and the flight after the flea leaves the
ground, with a downward acceleration. We treat these 2 parts separately and use a constant acceleration during the jump
(which would be the average acceleration in reality).
Fleas are amazing jumpers; they can jump several times their body height—something we cannot do.
We assume constant acceleration so we can use the equations in Table 2.4. The last of the three relates the three variables
we are concerned with in part (a): speed, distance (which we know), and acceleration (which we want).
2
2
(v y )f = (v y) i
2ay y
In part (b) we use the first equation in Table 2.4 because it relates the initial and final velocities and the acceleration
(which we know) with the time interval (which we want).
(vy )f =(vy )i ay t
Part (c) is about the phase of the jump after the flea reaches takeoff speed and leaves the ground. So now it is (v y )i , that is
1.0 m/s instead of (v y )f . And the acceleration is not the same as in part (a)—it is now g (with the positive direction up)
since we are ignoring air resistance. We do not know the time it takes the flea to reach maximum height, so we employ the
last equation in Table 2.4 again because we know everything in that equation except y.
Solve: (a) Use (v y )i = 0.0 m/s and rearrange the last equation in Table 2.4.
(vy ) 2f
2 y
ay
(1.0 m/s)2 1000 mm
2(0.50 mm)
1m
1000 m/s2
(b) Having learned the acceleration from part (a) we can now rearrange the first equation in Table 2.4 to find the time it
takes to reach takeoff speed. Again use (v y )i = 0.0 m/s.
(vy )f
t
ay
1.0 m/s
1000 m/s2
.0010 s
(c) This time (vy )f =0.0m/s as the flea reaches the top of its trajectory. Rearrange the last equation in Table 2.4
to get
y
vy
2ay
2
i
(1.0 m/s) 2
2( 9.8 m/s2 )
0.051 m 5.1 cm
Assess: Just over 5 cm is pretty good considering the size of a flea. It is about 10–20 times the size of a typical flea.
Check carefully to see that each answer ends up in the appropriate units.
The height of the flea at the top will round to 5.2 cm above the ground if you include the 0.050 cm during the initial
acceleration phase before the feet actually leave the ground.
2 dimensional problems
Leafy spurge disperses its seeds with an explosive ejection from the seed capsule and can
expel its seeds to distances up to 5 meters. The seeds have a mass of 3.04 mg each and are
accelerated through a distance of 2 mm. The seed pods are at the top of the plants which are
up to 1 meter tall.
Assuming the optimal 45º angle for the maximum distance, find the velocity with which the seed
is ejected to reach the 5 meter range. Ignore air resistance (which is likely a poor assumption).
Knowns:
yo = 1m xo = 0
θ = 45º
yf = 0
xf = 0
a = -g in flight
Because of the change in height, we cannot use the range formula from lab and the maximum
height is no longer reached at half the time for the entire flight; the symmetry of the trajectory is
destroyed. Instead, we will split the motion into vertical and horizontal components and solve
them independently.
Vertical:
Horizontal:
a=0, vxo = vo cos(45), xo = 0, xf = 5m
a = -g, vyo = vo sin(45), yo = 1m, yf = 0
y = yo + vyo t + ½ (-g) t2
= 1m + vo sin(45) t –g/2 t2
x = xo + vxo t = vo cos(45) t
vx = vo cos(45)
vy = vyo + ay t = vo sin(45) - gt
When the seed hits the ground at tf,
y = 0 = 1m + vo sin(45) tf –g/2 tf2
x = 5m = vo cos(45) tf solve for tf (simpler than quadratic)
Use tf from x equation, drop units:
0 = 1 +( vo/√2)(5√2/vo) – (g/2)(5√2/vo)2
= 1 + 5 – 25g/vo2
tf = 5m/ vo cos(45) = 5√2 / vo
25g/vo2 = 6
vo = 6.4 m/s
drop units to save space, equations consistent.
sin(45)=cos(45)=1/√2
vo2 = 25g/6 = 25(9.8)/6
t = 5√2 / vo = (5√2 [m]) / (6.4 [m/s]) = 1.1 s
We can now determine the acceleration of the seed as it is ejected from the pod.
During ejection, the velocity changes from 0 to 6.4 m/s over a distance of 2 mm. If we initially
ignore gravity, we can use
vf2 = vo2 +2ad
a = (vf2 - vo2 )/2d = 6.42/[2(.002)] = 1.02 x 104 m/s2
Comparing this acceleration to g, it appears that ignoring g was justified, since it is much smaller
than the value calculated.
Using Newton’s second law, we can calculate the force applied to the seed to eject it.
F = ma = (3.04 x 10-6 kg)(1.02 x 104 m/s2) = 3.10 x 10-2 kg•m/s2 = 3.10 x 10-2 N
Ramp problems:
Consider a block on a ramp
making an angle of 30º with the
horizontal.
Under normal conditions, the block will move along the ramp surface. The coordinates which
make the calculations the simplest are parallel to the ramp and perpendicular to it. This follows
the general strategy of
Choose one coordinate axis along the direction of the acceleration or motion
and the other perpendicular to that.
This means that the gravitational force needs to be split into components. I have given you
some geometry tools on the second exam data sheet, anticipating that the geometry you
learned in high school probably did not all stay in your ready – recall memory.
Applying
these tools to the gravitational force (mg)
gives the components in the diagram to the
right. The gravitational force can be
replaced by tits components, redrawing the
forces with the ramp direction horizontal
gives the diagram on the next page.
Unless the block is sinking into the ramp or jumping up
off it, the acceleration perpendicular to the ramp is zero,
as is the velocity. The acceleration along the ramp is
downward or to the right in the diagram.
Force balance:
Perpendicular FN - mg cos(30) = 0 or FN = mg cos(30)
Along ramp: Fnet = mg sin(30) – f =ma
(a) If the ramp is frictionless, then Fnet = mg sin(30) and the acceleration of the block is given
by
a = g sin(30) = g/2.
This gives
v = vo + a t = vo + (g/2)t
and
d = do + vo t + ½ a t2 = do + vo t + ½ (g/2) t2
For displacement, take do = 0
= vo t + ½ (g/2) t2
(b) If there is friction, the friction force can be represented by f = μ FN where μ is the coefficient
of friction. Let’s see what we can determine from this and our force balance equations above.
f = μ FN and, from the perpendicular force balance,
FN = mg cos(30)
giving f = μ mg cos(30)
Then
Fnet = mg sin(30) – f = mg sin(30) – μ mg cos(30)
= mg [sin(30) – μ cos(30)] = ma
or
a = g [sin(30) – μ cos(30)].
Does this suggest a way to measure the coefficient of static friction (μs) when the block is
standing still on the ramp?
1. A student brings in her physics homework, dragging it across campus in a
wooden box. She is pulling with a force of 450 N on a rope, which inclined at
38 degrees to the horizontal. The ground exerts a horizontal force of 125N
that opposes the motion. Calculate the acceleration of the crate if (a) it
mass is 310 kg and (b) its weight is 310 N.
First, the crate will move
(be accelerated) in the
horizontal direction. Use coordinate axes which are horizontal and vertical.
Resolve the 450N pull into horizontal and vertical components.
The simplified force diagram is now:
Vertical: FN + 450sin(38) = mg
Horizontal: 450cos(38) – 125 = ma
a) m = 350 kg:
a
b) wt = 350N = mg
(450cos(38) 125) N
350kg
0.66m / s 2
m = 350/g = 35.7 kg
a
(450cos(38) 125) N
35.7kg
6.43m / s 2
You can also calculate the Normal force from the ground:
a) FN + 450sin(38) = mg = (350 kg)(9.8m/s2) = 630 N
FN = 630 - 450sin(38) = 353 N
b) FN + 450sin(38) = mg = 350 N
FN = 350 - 450sin(38) = 73 N
2.
m1 is 3.70 kg on a frictionless inclined plane. Pulley is massless, and
frictionless. m2 is 2.30 kg.
(a) What is magnitude of acceleration of each block?
(b) What is direction of acceleration of hanging block,
m2? (c) What is value of the tension in the cord?
Show force diagrams for each mass.
Key ideas for solving text problems:
(1) The tension in a single piece of rope is
same anywhere in the rope and at both
ends, as long as it is massless or nearly so.
(2) The only effect of a massless,
frictionless pulley is to change the direction
the force, it does not affect the size of the
force.
the
of
(3) Objects connected by string will have the same acceleration. (What would happen if this
were not true?)
From the diagram of the block on the ramp, the simplest axes are going to be along the ramp
and perpendicular to it (you only need to break up one vector that way). This gives the following
diagram with the components of the weight (force of gravity) included:
For this block,
FN = m1g cos(θ) from the vertical balance
and
T – m1g sin(θ) = m1a along the ramp.
For the hanging block, m2g – T = m2a
For connected objects like this, it is frequently fruitful to look at the
entire set of connected objects as one total mass and set up the
forces for that.
Here, that gives m2g – m1g sin(θ) = (m1 + m2)a Which will allow
us to solve for the acceleration a.
Let’s look at the forces along the ramp and for the hanging mass.
T – m1g sin(θ) = m1a
m2g – T = m2a
We want to solve for tension T and acceleration a, so let me rearrange these to give
T - m1a = m1g sin(θ)
T + m2a = m2g
Subtracting the first equation from the second leaves the equation above for the two masses
together:
m2a + m1a = m2g – m1g sin(θ) (The T’s subtract out).
Once we solve this for a, we can use that value in either equation above to find the tension T.
If it is helpful, there is a not on the web site describing various ways to work with two equations
in two unknowns.
Some miscelaneous force diagrams: