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A particle Ptravels with constant speed on a circle of radiusr=3.00m(Fig. 4-56) and completes one revolution in 20.0s. The particle passes through Oat time t= 0. State the following vectors in magnitude angle notation (angle relative to the positive direction of x).With respect to 0, find the particle’s position vectorat the times tof (a) 5.00s, (b)7.50s, and (c)10.0s. (d) For the 5.00sinterval from the end of the fifth second to the end of the tenth second, find the particle’s displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

Short Answer

Expert verified

(a) The position vector of particle P after 5secis 4.2m,45°.

(b) The position vector of particle P after7.5sec is 5.5m,68°.

(c) The position vector of particle P after10secis6m,90°.

(d) The position vector of particle P from the end of the fifth second to the end of the tenth-second interval is4.2m,135°.

(e) For the above interval, the particle’s average velocity is0.85m/s,135°.

(f) The particle’s velocity at the beginning isrole="math" localid="1657014983012" 0.94m/s,90°.

(g) The particle’s velocity at the end is0.94m/s,180°.

(h) The particle’s acceleration at the beginning is0.30m/s2,180°.

(i) The particle’s acceleration at the end of the interval is 0.30m/s2,270°.

Step by step solution

01

Given data

1) The radius of the circle is r=3.00m.

2) The time period of revolution is T=20s.

02

Concept of radius vector and periodic motion

The magnitude of radius vector will be same but the direction will keep changing as it rotates. As the direction keeps changing, we will have different values of velocity and acceleration at different time. We can use trigonometric relation and find magnitude angle notation. By using expression of average velocity, we can find coordinates of the radius vector, velocity, and acceleration at different time.

03

(a) The position vector at the time t=5.0s

At the timet=5.0s, the particle has traveled a fraction of the position vector.

tT=520=14

The angle will be for that time is

Φ=14x360=90°

This angle is measured from the vertical. Then the position coordinate from the figure X=3.0mandy=3.0mrelative to the coordinate origin.

The magnitude of R is,

R=x2+y2=3.0m2+3.0m2=4.2m

The angle is,

θ=tan-1yx=tan-13.003.00=45°

In magnitude the displacement vector and angle notation is4.2m,45°

04

(b) The position vector at time t=7.5s

The particle has traveled a fraction of7.520=38revolution around the circle. The angle of the particle at that instant,

Φ=36x360=135o

This is the angle measured from the vertical.

From the trigonometric relation,

x=rsinΦ=3.0msin135°=2.1m

y=r-rcosΦ=3.0m-3.0mcos135=5.1m

The magnitude of R is,

R=x2+y2=2.1m2+5.1m2=5.5m

The angle is,

θ=tan-15.121=68°

In magnitude angle notation, the displacement vector is5.5m,68°

05

(c) The position vector at time t=10s

The particle has traveled a fraction of1020=12revolution around the circle. The angle of the particle at that instant,

Φ=12x360=180°

This is the angle measured from the vertical.

From the figure,x=0mandy=6.0mare position coordinates relative to the coordinate origin.

In magnitude angle notation, the displacement vector is6.0m,90°

06

(d) Calculate the displacement vector from the start of fifth and at the end of tenth second.

We can subtract the position vector in part (a) from the position vector in (c)

6.0,90°-4.2,45°=4.2,135°

In unit vector notation,

R=0-3.0mi^+6.0m-3.0mj^=3.0mi^+3.0mj^

The magnitude of the displacement vector is,

R=3m2+3m2=4.2m

Therefore, in magnitude angle notation, the displacement vector is4.2m,135°

07

(e) Calculate the average velocity from the start of fifth and at the end of tenth second.

The time interval is t=5.0s. By using the expression of average velocity expression,

Vavg=Rt=3.0mi^+3.0mj^0.5=-0.60m/si^+0.60m/sj^

The magnitude of the velocity vector,

Vavg=-0.60m/s2+0.60m/s2=0.85m/s

Therefore, in magnitude angle notation, the velocity vector is0.85m/s,135°

08

(f) Calculate the velocity at the beginning

The speed of the particle is calculated by dividing the circumference of the circle with the time of revolution.

V=2πrT

Substitute the 3mforr, and 20sfor T in the above equation.

=2π×3.00m20.00s=0.942m/s

The velocity vector is tangent to the circle at its 3 O’ clock position at which vis vertical. Therefore, the velocity at the beginning is0.94m/s,90°.

09

(g) Calculate the velocity at the end

The velocity vector is tangent to the circle at its 12 O’ clock position at whichvis vertical. The speed will be same as calculated in part (f).

Therefore, the velocity at the end is0.94m/s,180°

10

(h) Calculate the acceleration at the beginning

At the beginning, the magnitude of the acceleration is calculated as,

a=V2r

Substitute the 3mforr, and 0.94m/s2for Vin the above equation.

=0.94m/s23.00m=0.30m/s2

The acceleration is horizontal (from part a) and directed towards the center of the circle. Therefore, in magnitude angle notation, the acceleration at the beginning is0.30m/s2,180°

11

i) Calculate the acceleration at the end

Similarly, at the end, the acceleration is,

a=V2r

Substitute the 3mforr, and 0.94m/s2for Vin the above equation.

=0.94m/s23.00m=0.30m/s2

At this instant, acceleration is vertical (from part (c)) and directed towards the center. Therefore, in magnitude angle notation, the acceleration at the beginning is 0.30m/s2,270°.

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