A small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. Which set of vectors gives the di
rection of the velocity v with arrow, acceleration a with arrow of the cylinder, and the net force ΣF with arrow that acts on it?
(A) 3
(B) 4
(C) 1
(D) 2
(A) 3
(B) 4
(C) 1
(D) 2
1 answer:
The question is missing the diagram. Also, the choices must have pictorial representation. So, I have attached the missing diagram and the pictorial representation of the vectors.
Answer:
The correct representation is attached below. Force and acceleration will be towards the center of rotation while the velocity will be along the tangent to the circular motion. <u>Option (D).</u>
Explanation:
From the figure, we can conclude the following points:
1. The cylinder is under a uniform circular motion as the circular table is moving at constant speed.
2. For a circular motion, velocity acts along the tangent to the circular path.
3. For a circular motion, centripetal force acts on the body that causes it move around a circular path.
4 The direction of the centripetal force is radially inward towards the center of rotation.
5. The centripetal force causes a centripetal acceleration acting on the body.
6. From Newton's second law, the net acceleration of a body is in the same direction as that of the net force acting on it. So, centripetal acceleration also acts in the radially inward direction.
Therefore, from the above conclusions, it is clear that velocity will act in the horizontal direction at the given instance of time and force and acceleration will act vertically down for the given instance.
This is shown in the picture below. The option (D).
You might be interested in
Explanation:
Attachments of the solution are below.
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
