Question

The particle has a mass of 0.8 kg and is confined to move along the smooth...

The particle has a mass of 0.8 kg and is confined to move along the smooth vertical slot due to the rotation of the arm OA. The rod is rotating with a constant angular velocity θ  = 2 rad/s. Assume the particle contacts only one side of the slot at any instant.(Figure 1).

Part A

Determine the magnitude of the force of the rod on the particle when θ  = 30°. Express your answer to three significant figures and include the appropriate units.

Part B

Determine the magnitude of the normal force of the slot on the particle when θ  = 30°. Express your answer to three significant figures and include the appropriate units.

The particle has a mass of 0.8 kg a

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Answer #1
Concepts and reason

Free body diagram:

Free body diagram is the representation of all the forces acting on the body.

Newton’s 2nd law:

It states that, the rate of change in momentum is directly proportional to the force applied.

Fundamentals

Expression for the components of the acceleration of the particle in terms of polar coordinates r and are:

Radial acceleration, a, r-r

Tangential acceleration, a, r-r

Equations of motion in terms of polar coordinates r and are:

Along radial direction, ΣΕ-ma,

Along tangential direction, ΣΕ-ma

From trigonometry,

cos00.5
r

0.5
r=
cose
=0.5sec 0
…… (1)

-0.5 sec30°
=0.5774 m

Differentiate equation (1) with respect to time to get the velocity.

dr
d
(0.5sec0)
dt dt
r = 0.50sec0 tan 0
…… (2)

0.5(2)sec (30°)tan (30°)
=0.6667 m/s

Differentiate equation (2) with respect to time to get the acceleration.

dr
d
(0.50sec0 tan e)
0.5[0sec tan 0+0sec (0sec* 0)+0tan(0sec0 tan o)
dt dt
r = 0.5| 0sec0 tan0+0° (sec0)+(0 sec0tan26
= 0.5 0sec 30° tan 30° +(2 (see 30)+(2jsec30° tan 30)
=0.5[6.1584+1.5396
=3.849 m/s2

Calculate the radial acceleration of the particle using the following equation:

a, r-r0
3.849-0.5774 (2)
1.539 m/s

Calculate the tangential acceleration of the particle using the following equation:

a, r0+2r0
(0.5774)(0)+2(0.6667) (2)
2.667 m/s2

Draw the free body diagram of the particle.

t
та,
r
F.
та,
slot
F
mg
rod

Consider the force equilibrium along the radial direction.

ΣΕ-ma
Fot cos0-mg sin 0 = ma,
= ma,

F COs (30)(0.8)(9.81)sin (30°) (0.8) (1.539)
0.866F-3.924 1.2312
slot
slot
F 5.95 N
slot

Consider the force equilibrium along the tangential direction.

XF-ma,
Fndt Sin0-mg cos0 = ma,
Fnd-(5.95)sin (30°)(0.8)(9.81)cos (30)(0.8)(2.667)
F-2.975-6.8 = 2.1336
slot
rod
rod
rod
F 11.

Ans: Part b

The magnitude of the normal force of the slot on the particle is 5.95 N
.

Part a

The magnitude of the force of the rod on the particle is 11.9 N
.

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