Question

6. An initially stationary wheel has angular acceleration 2.6 rad/s2. At t = 5 s, find (a) the angular velocity in rad/s, (b) the linear velocity in m/s of a point 0.4 m from the center, (c) how many radians and (d) how many revolutions the wheel has rotated through.

7. An oxygen molecule consists of two atoms, each of mass 2.7x10 ^-26 kg, separated by distance 1 .2x10^10 m. Calculate the moment of inertia about an axis perpendicular to the line joining the atoms, midway between them.

Answer 1

6) initial angular velocity wi = 0 rad/s

t = 5 sec

alpha = 2.6 rad/s^2

wf = ?

use wf = wi+(alpha*t) = 0+(2.6*5*5) = 65 rad/s

B) linear velocity is v = r*w = 0.4*65 = 26 m/s

C) theta = (wi*t)+(0.5*alpha*t^2) = 0 + (0.5*2.6*5*5) = 32.5 rad

D) 32.5/(2*pi) = 32.5/(6.284) = 5.17 rev = 5 rev

7) moment of inertia is I = 2*m*r^2 = 2*2.7*10^-26*(1.2*10^10/2)^2 = 1.944*10^-6 kg-m^2

Answer 2

SOLUITION :

6.

a.

Angular acceleration = alpha = 2.6 rad/sec^2

Initial angular velocity = 0

Angular velocity, omega,  after 5 secs 

= 2.6 * 5 

= 13 rad/sec (ANSWER)

b.

Linear velocity of a point 0.4 m from centre (after 5 secs) :

= r * omega

= 0.4 * 13

= 5.2 m/sec (ANSWER).

c.

Average angular velocity , omega average, during 5 sec interval 

= (0 + 13) / 2 

= 6.5 rad/sec

Radians wheel moved 

= average angular velocity * time

= 6.5 * 5

= 32.5 radians (ANSWER).

d.

Revolutions wheel rotated in 5 sec duration 

= 32.5 / (2 * π)

= 5.17 revolutions (ANSWER).

7.

M. I. 

= m1 * r^2 + m2 r^2 

= 2 m r^2  (since m1 = m2)

= 2 * 2.7*10^(-26) * ((1.2 * 10^(10) / 2)^2

= 1.944 * 10^(-6) kg-m^2 (ANSWER).