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Question

http x C https:// myasucourses.asu.edu 1/courses/2015SummerC-T-MAE212-42430/SampleExam2.pdf 4. A particle is travelling on a circular track of radius 4m. Particle started on the x-axis and its speed is proportional to the distance travelled along the arc of the circle. So we have v FS. When the distance travelled is s meters. a. What is the magnitude of the acceleration of the particle? b. What is the x component of the velocity of the particle? Will the particle ever be able to complete a full circle while having its speed proportional to the distance travelled? The coefficient of friction between the particle and the track is 1.2 Hint: the particle will start to slip off the track when the magnitude of its total acceleration becomes greater than ug where g 39.8 rm

Explanation / Answer

here,

particel is moving in a circular path,

radius of circular track r=4m

given that, v=s

===>

s=pi=3.14 m

v=3.14 m/sec^2

a)

acceleration a=sqrt[(a_rad)^2 +(a_tan)^2]

here,

a_rad=v^2/r

=pi^2/r

=3.14^2/4

=2.46 m/sec^2

and

a_tan=v/t=pi=3.14 m/sec^2

now,

a=sqrt(2.46^2+3.14^2)

=3.98 m/sec^2

b)

vx=-v*cos(theta)

here,

theta=s/r=3.14/4 =0.785 rad

=44.9 degrees

vx=-v*cos(44.9)

=3.14*cos(44.9)

=2.22 m/sec

c)

here,

co-efficent of friction u=1.2

acceleration of particle a' =u*g

a'=1.2*9.8

a'=11.76 m/sec^2

now,

to complete one circle,

distance travelled s=2*pi*r

=2*3.14*4

=25.12 m

====>v=25.12 m/sec^2

and

a=v^2/r

=25.12^2/4

=157.75 m/sec^2

hence to complete one circle accelertaion must be a=157.75 m/sec^2

and which is greater than a'=11.76 m/sec^2

hence, particle slip off the track and with these condition(v=s) particle won't complete the circle

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