Question

A wheel 2.40 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.40 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. What is the tangential speed?
B. Total acceleration
C. Angular position of point P.

Answer 1

Step-by-step explanation:

Radius of wheel = 1.2 m

A )

To know angular speed after t sec , we use the formula

ω = ω₀ + α t , where ω₀ is initial velocity , α is angular acceleration

ω = 0 + 4.4 x 2

= 8.8 rad / s

v= ωR , v is tangential speed , ω is angular speed , R is radius of wheel .

= 8.8 x 1.2 = 10.56 m /s

B )

radial acceleration

Ar = v² / R

= 10.56² / 1.2

= 92.93 m /s²

Tangential acceleration

At = angular acceleration x radius

= 4.4 x 1.2 = 5.28 m /s²

Total acceleration

= √ ( At² + Ar² )

=√ (5.28² +92.93²)

= 93 m /s²

C )

θ = ωt + 1/2 α t² where θ is angular position after time t .

= 0 + .5 x 4.4 x 2²

= 8.8 rad

= 180x 8.8/ 3.14 = 504.45 degree

initial position = 57.3°

final position = 504 .45 + 57.3

= 561.75 °

= 561.75 - 360

= 201.75 ° .

Position of radius vector of point P will be at angle of 201.75 from horizontal axis .