Question

A wheel 2.45 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.30 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.
a. the angular speed of the wheel and, for point P
b. the tangential speed.
c. the total acceleration.
d. the angular position.

Answer 1

Answer:

Explanation:

Radius of wheel R = 1.225 m

For angular motion of wheel

ω =   ω ₀ + α t

= 0 + 4.3 x 2

= 8.6 rad / s

This is angular speed of wheel and point P .

b )

Tangential speed = ωR

8.6 x 1.225

= 10.535 m / s

c )

radial acceleration

a_r = v² / r

= 10.535² / 1.225

= 90.6 m / s²

tangential acceleration = radius x angular acceleration

a_t = 4.3 x 1.225

= 5.2675

Total acceleration = √ 90.6² + 5.2675²

=  √ 8208.36 + 27.7465

= 90.75 m/s²

d ) angle of rotation

= 1/2 α t²

= .5 x 4.3 x 4

= 8.6 radian

= (8.6/3.14) x 180

= 499 degree

= 499 + 57.3

= 556.3

556.3 - 360

= 196.3 degree

Point p will rotate by 196.3 degree