A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of t
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Answer:
Force = 186 N
Explanation:
Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :
Where is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector),
is the applied force.
The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that (the magnitud of the position vector) is 35cm.
Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :
Now using the torque formula:
Where is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector
, thus :
so the force is approximately 186 N.
Answer:
The average speed of the fielder is 5.24 m/s
Explanation:
The position vector of the ball after it was hit can be calculated using the following equation:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem.
When the ball is caught, its position vector will be (see r1 in the figure):
r1 = (r1x, 0.873 m)
Then, using the equation of the position vector written above:
r1x = x0 + v0 · t · cos α
0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²
Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:
r1x = v0 · t · cos α
0.873 m = v0 · t · sin α + 1/2 · g · t²
Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:
0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²
0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²
Solving the quadratic equation:
t = 0.03 s and t = 5.72 s.
It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.
With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:
r1x = v0 · t · cos α
r1x = 37.2 m/s · 5.72 s · cos 49.3°
r1x = 1.39 × 10² m
The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.
The average velocity is calculated as the traveled distance over time, then:
average velocity = treveled distance / elapsed time
average velocity = 30.0 m / 5.72 s = 5.24 m/s
Answer:
a) T = 1.69 s, b) k = 0.825 N / m, c) v = 1.46 feet/s, d) a = 5.41 ft / s²,
e) v = - 1,319 ft / s, a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J
Explanation:
In a mass-spring system with simple harmonic motion, the angular velocity is
w =
a) find the period
angular velocity, frequency, and period are related
w = 2π f = 2π / T
f = 1 / T
T = 1 / f
T = 1 / 0.59
T = 1.69 s
b) the spring constant
w = 2π f
w = 2π 0.59
w = 3.70 rad / s
w² = k / m
k = w² m
k = 3.70² 0.060
k = 0.825 N / m
c) the maximum speed
simple harmonic movement is described by the expression
x = A cos (wt + Ф)
speed is defined by
v =
v = -A w sin (wt + fi)
the speed is maximum when the cosine is ± 1
v = A w
v = 0.394 3.70
v = 1.46 feet/s
d) maximum acceleration
a =
a = - A w² cos wt + fi
the acceleration is maximum when the cosine is ±1
a = A w²
a = 0.394 3.70²
a = 5.41 ft / s²
e) velocity and acceleration for x = 6 cm
let's reduce the cm to feet
x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot
Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s
let's use the expression for the velocity
v = -A w sin (0 + Фi)
0 = - A w sin Ф
so sin Ф = 0 which implies that Фi = 0
the equation of motion is
x = A cos wt
x = 0.394 cos 3.70t
we substitute
0.1969 = 0.394 cos 370t
3.70 t = cos⁻¹ (0.1969 / 0.394)
let's not forget that the angle is in radians
3.70, t = 1.047
t = 1.047 / 3.70
t = 0.2826 s
we substitute this time in the equation for velocity and acceleration
v = - Aw sin wt
v = - 0.394 3.70 sin 3.70 0.2826
v = - 1,319 ft / s
a = - A w² cos wt
a = - 0.394 3.70² cos 3.70 0.2826
a = - 2.70 ft / s²
f) the kinetic and potential energy at this point
K = ½ m v²
let's slow down to the SI system
v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s
K = ½ 0.060 0.402²
K = 4.8 10⁻³ J
U = ½ k x²
U = ½ 0.825 0.06²
U = 1.49 10⁻³ J