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A 0.2‐kg particle P is constrained to move along the vertical‐plane circular slot of radius r = 0.5 m and is confined to the slot of arm OA, which rotates about a horizontal axis through O with a constant angular rate Ω = 3 rad/s. For the instant when β = 20°, determine the force N exerted on the particle by the circular constraint and the force R exerted on it by the slotted arm.

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abidhussain7972

Answer:

● The force N exerted on the particle by the circular constraint= -2.885 N

● The force R exerted on it by the slotted arm= 1.599 N

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

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jayilych4real

The force N exerted on the particle by the circular constraint= -2.885 N

The force R exerted on it by the slotted arm= 1.599 N

Calculations and Parameters:

Using the free body diagram,

To find the radius,

cos [tex]\beta[/tex]= R/ 2 x 0.5

R= 2(0.5 cos 20°)

= 0.9396m.

Differentiate with respect to time:

R= -sin20° x 3

= -1.026m/s.

Differentiate also with respect to time:

R= 0-cos20° x (3)^2

= -8.457m/s^2.

After we calculate along the normal and tangential direction, we would get:

  • -16.9134 m/s^2
  • -6. 156 m/s^2

Thus, the dynamic equilibrium condition along r-direction will be:

0.9396N= -3.3826 + 0.671

= -2.885N.

Hence, the dynamic equilibrium condition along θ-direction will be:

R= -1.2312 + 2.83

= 1.599N.

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