Answer:
The initial angular speed of the CD is equal to 14.73 rad/s.
Explanation:
Given that,
Angular displacement,
Final angular speed,
The acceleration of the CD,
We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,
Put all the values,
So, the initial angular speed of the CD is equal to 14.73 rad/s.
Answer:
a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,
c) F = 281.70 N, d) θ = 338.58º , e) a = 3,588 m / s² , f) θ = 201.45º
Explanation:
For this exercise we will use Newton's second law on each axis
X axis
-Tₓ = m aₓ
Y Axisy
–W = m a_{y}
Let's use trigonometry to find the components of force
sin 21.5 = Tₓ / T
cos 21.5 = T_{y} / T
Tₓ = T sin 21.5
T_{y} = T cos 21.5
Tₓ = 768 sin 21.5 = 281.47 N
T_{y} = 768 cos 21.5 = 714.56 N
a) the force of the rope on Tarzan is
T = (281.47 i ^ + 714.56 j ^) N
b) The net force is the subtraction of the tension minus the weight of Tarzan
Y Axis F_net = 714.56 - 825 = -110.44 N
F_net = (281.47 i ^ - 110.44 j ^) N
c) Let's use Pythagoras' theorem
F = √ (Fₓ² + T_{y}²)
F = √ (281.47² + 110.44²)
F = 281.70 N
d) Let's use trigonometry
tan θ = F_{y} / Fₓ
θ = tan⁻¹ F_{y} / Fₓ
θ = tan⁻¹ (-110.44 / 281.47)
θ = -21.42º
This angle is average clockwise, for counterclockwise measurement
θ = 360 - 21.42
θ = 338.58º
Acceleration
X axis
Tₓ = m aₓ
aₓ = Tₓ / m
The mass of Tarzan is
m = W / g
m = 825 / 9.8 = 84.18 kg
aₓ = 281.47 / 84.18
aₓ = -3.34 m / s2
Y Axis
T_{y}-W = m a_{y}
a_{y} = (T_{y} -W) / m
a_{y} = (714.56-825) / 84.18
a_{y} = - 1,312 m / s²
Acceleration Module
a = √ aₓ² + a_{y}²
a = √ (3.34² +1.312²)
a = 3,588 m / s²
The angle
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-1312 / -3.34)
θ = 21.45º
Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis
θ = 180 + 21.45
θ = 201.45º
Answer:
approximately 30 degrees
Explanation:
If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):
By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:
which tells us that the closer answer shown is