Explanation:
- vector r lies on z- axis
- J is tilted at angle Ψ
- Orient x-axis such that w lies in x-z plane
Given:
Vector potential
Where, K = б*v ; r* = sqrt (R^2 + r^2 -2R*r*cos(θ')) ; da' = R^2*sin(θ')*dθ'dΦ'
Solution:
- Velocity of v point a point r' in a rotating rigid body is given by:
v = w x r' =
- where a = Ψ and b' = θ' and c' = Φ'
v = R*w [-(cos Ψ *sin θ' *sin Φ') x + (cos Ψ *sin θ' *cos Φ' - sin Ψ * cos θ') y
+ (cos Ψ *sin θ' *sin Φ') z ]
- Notice that terms like sin Φ' and cos Φ' contribute to zero:
- Hence,
- Evaluate integral u = cos (b')
- From we can determine two cases when r > R and r < R
Hence,
r < R
r > R
- Reverting back to original coordinate system given in figure 5.45:
r < R
r > R
Where, b = θ and c = direction along Φ.
Hence, A ( r , θ , Φ )
The amount of force required to accelerate the given mass of the wagon is 129 Newtons.
<h3>What is force?</h3>
A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
From Newton's Second Law, force is expressed as;
F = m × a
Where is mass of object and a is the acceleration.
Given the data in the question;
- Mass of the rock m = 8.6kg
F = 8.6kg × 15m/s²
F = 129kgm/s²
F = 129N
Therefore the amount of force required to accelerate the given mass of the wagon is 129 Newtons.
Learn more about force here: brainly.com/question/27196358
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Answer:
Speed of a child on the rim is 4.25 m/s.
Explanation:
Given;
diameter of the merry-go-round, d = 5.0 m
period of the motion, t = 3.7 s
one complete rotation of the merry-go-round = πd
one complete rotation of the merry-go-round = π(5) = 15.71 m
Speed is given as distance / time
speed of a child on the rim = 15.71 / 3.7
speed of a child on the rim = 4.25 m/s
Therefore, speed of a child on the rim is 4.25 m/s.
Answer:
The drawing of the free body diagram is in the attachment.
Explanation:
In order to draw a<em> free body diagram</em>, you have to include in the drawing all the acting forces in the box for the given situation.
The acting forces are:
-The normal force (N), which is the contact force that the inclined plane applies to the box. This force is perpendicular to the contact surface.
-The weight force (W), which is due to the force of gravity on the box. The direction of the weight force is vertical (perpendicular to the ground)
-The friction force (Fs), which in this case is the static friction force because the box is in rest. The direction of this force is opposite to the direction of the movement.
Therefore, you have to draw the inclined plane, the box and the acting forces (See the attachment)