In this example we will apply the equations for constant angular acceleration to

Question

In this example we will apply the equations for constant angular acceleration to the simple case of a rotating bicycle wheel. The angular velocity of the rear wheel of a stationary exercise bike is 0=4.00rad/s at time t=0, and its angular acceleration is constant and equal to =2.50rad/s2. A particular spoke coincides with the +x axis at time t=0 (Figure 1) . What angle does this spoke make with the +x axis at time t=3.00s? What is the wheel's angular velocity at this time?

SOLUTION

SET UP We take =0 at the postion where the chosen spoke is horizontal, and we take the counterclockwise rotation to be positive. The initial conditions are 0=0, 0=4.00rad/s, and =2.50rad/s2.

SOLVE We can use the equations for constant angular acceleration to find and at any time. The angle is given as a function of time

===0+0t+12t20+(4.00rad/s)(3.00s)+12(2.50rad/s2)(3.00s)223.3rad=(23.3rad)1rev2rad=3.70rev

The wheel has turned through three complete revolutions plus an additional 0.70 rev , or(0.70rev)(2rad/rev)=4.40rad=252. Thus, the chosen spoke is at an angle of 252 with the +x axis.

The wheel's angular velocity is given as a fuction of time. At a time t=3.00s,

==0+t4.00rad/s+(2.50rad/s2)(3.00s)=11.5rad/s

Alternatively,

2===02+2(0)(4.00rad/s)2+2(2.50rad/s2)(23.3rad)=132rad2/s211.5rad/s=1.83rev/s

REFLECT Note the step-by-step similarity between these equations and the kinematic relationships for straight-line motion.

Part A - Practice Problem:

How much longer will it take before the reference spoke is once again aligned with the +x axis?

Express your answer in seconds to three significant figures.

SubmitMy AnswersGive Up

Part B - Practice Problem:

What will the angular speed of the wheel be at this time?

Express your answer in radians per second to three significant figures.

SubmitMy AnswersGive Up

In this example we will apply the equations for constant angular acceleration to the simple case of a rotating bicycle wheel. The angular velocity of the rear wheel of a stationary exercise bike is 0=4.00rad/s at time t=0, and its angular acceleration is constant and equal to =2.50rad/s2. A particular spoke coincides with the +x axis at time t=0 (Figure 1) . What angle does this spoke make with the +x axis at time t=3.00s? What is the wheel's angular velocity at this time?

SOLUTION

SET UP We take =0 at the postion where the chosen spoke is horizontal, and we take the counterclockwise rotation to be positive. The initial conditions are 0=0, 0=4.00rad/s, and =2.50rad/s2.

SOLVE We can use the equations for constant angular acceleration to find and at any time. The angle is given as a function of time

===0+0t+12t20+(4.00rad/s)(3.00s)+12(2.50rad/s2)(3.00s)223.3rad=(23.3rad)1rev2rad=3.70rev

The wheel has turned through three complete revolutions plus an additional 0.70 rev , or(0.70rev)(2rad/rev)=4.40rad=252. Thus, the chosen spoke is at an angle of 252 with the +x axis.

The wheel's angular velocity is given as a fuction of time. At a time t=3.00s,

==0+t4.00rad/s+(2.50rad/s2)(3.00s)=11.5rad/s

Alternatively,

2===02+2(0)(4.00rad/s)2+2(2.50rad/s2)(23.3rad)=132rad2/s211.5rad/s=1.83rev/s

REFLECT Note the step-by-step similarity between these equations and the kinematic relationships for straight-line motion.

Part A - Practice Problem:

How much longer will it take before the reference spoke is once again aligned with the +x axis?

Express your answer in seconds to three significant figures.

  s  

SubmitMy AnswersGive Up

Part B - Practice Problem:

What will the angular speed of the wheel be at this time?

Express your answer in radians per second to three significant figures.

  rad/s  

SubmitMy AnswersGive Up

Explanation / Answer

Part A) angular diplacement is theta = 2*pi = 6.284 rad

so using

theta = (wi*t)+(0.5*alpha*t^2)

6.284 = (4*t)+(0.5*2.5*t^2)

t = 1.15 sec

B) angular speed is W = wi+(alpha*t) = 4+(2.5*1.15) = 6.875 rad/s

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