15. %u2013 /2 points CJ9 8.P.043. My Notes | 15. %u2013 /2 points CJ9 8.P.043. M

Question

15.%u2013/2 pointsCJ9 8.P.043.My Notes  | 15.%u2013/2 pointsCJ9 8.P.043.My Notes  | Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)
rad/s

(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?
rad/s2
16.%u2013/1 pointsCJ9 8.P.053.My Notes  | 16.%u2013/1 pointsCJ9 8.P.053.My Notes  | Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used A motorcycle accelerates uniformly from rest and reaches a linear speed of 20.4 m/s in a time of 8.91 s. The radius of each tire is 0.282 m. What is the magnitude of the angular acceleration of each tire?
rad/s2
17.%u2013/2 pointsCJ9 8.P.054.My Notes  | 17.%u2013/2 pointsCJ9 8.P.054.My Notes  | Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used (a) Determine the angular speed of the wheel. (Assume that there is no slipping of the surfaces in contact during the rolling motion.)
rad/s

(b) Relative to the axle, what is the tangential speed of a point located 0.175 m from the axle?
m/s
18.%u2013/1 pointsCJ9 8.P.061.My Notes  | 18.%u2013/1 pointsCJ9 8.P.061.My Notes  | Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used Question Part Points Submissions Used

Explanation / Answer

1)a)conserving mechanical energy,

0.5Iw^2=mg(L/2)

or 0.5*(1/3)*w^2*1.762^2=9.8*0.5*1.762

or w=4.08 rad/s

b)balancing torque,

mgL/2=I*alpha

or 9.8*1.762*0.5=(1/3)*1.762^2*x

or x=8.34 rad/s^2

2)linear acceleration=20.4/8.91

=2.289 m/s

so x=v/r

= 2.289/0.282

=8.12 rad/s^2

3)a)w=v/r

=14/0.43

=32.55 rad/s

b)v=wr

=32.55*0.175

=5.697 m/s

4)distance covered is same.so,

n*r=constant

so,

1.2*446=0.34*x

or x=1574.12 revolutions

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