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

Answer 2

(a) The angular speed of the wheel at point P is 8.6 rad/s.

(b) The tangential speed of the wheel is 10.54 m/s.

(c) The total acceleration of the wheel is 90.8 m/s².

(d) The angular position of the wheel is 87 ⁰.

The given parameters;

• diameter of the wheel, d = 2.45 m
• radius of the wheel, r = 1.225 m
• angular acceleration of the wheel, α = 4.3 rad/s²
• angular displacement of the wheel, θ = 57.3⁰
• time of motion, t = 2.0 s

The angular speed of the wheel at point P is calculated as follows;

$\omega_f = \omega _i + \alpha t\\\\\omega _f = 0 + 4.3 \times 2\\\\\omega _f = 8.6 \ rad/s$

The tangential speed of the wheel is calculated as follows;

$v = \omega _f r\\\\v = 8.6 \times 1.225 \\\\v = 10.54 \ m/s$

The centripetal acceleration of the wheel is calculated as follows;

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(10.54)^2}{1.225} \\\\a_c = 90.69 \ m/s^2$

The total acceleration of the wheel is calculated as follows;

$a_t = \sqrt{a_c^2 + a_r^2} \\\\a_t = \sqrt{90.69^2 + 4.3^2} \\\\a_t = 90.8 \ m/s^2$

The angular position is calculated as follows;

$\theta = tan^{-1} (\frac{a_c}{a_r} )\\\\\theta = tan^{-1} (\frac{90.69}{4.3} )\\\\\theta = 87 \ ^0$

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