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Chapter 11: Problem 29

Coupled Wheels In Fig. \(11-25\). wheel \(A\) of radius \(r_{A}=10 \mathrm{~cm}\) is coupled by belt \(B\) to wheel \(C\) of radius \(r_{C}=25 \mathrm{~cm}\). The rotational speed of wheel \(A\) is increased from rest at a constant rate of \(1.6 \mathrm{rad} / \mathrm{s}^{2}\) Find the time for wheel \(C\) to reach a rotational speed of \(100 \mathrm{rev} / \mathrm{min}\),assuming the belt does not slip. (Hint: If the belt does not slip, the translational speeds at the rims of the two wheels must be equal.)

Short Answer

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16.36 seconds

Step by step solution

01

Determine the relationship between rotational speeds

If the belt does not slip, the linear (tangential) velocities at the rims of wheel A and wheel C must be equal. The formula for tangential velocity is given by: \[v = r \times \text{angular velocity} (\omega)\]Thus, for wheels A and C, we have:\[r_A \times \omega_A = r_C \times \omega_C\]
02

Calculate the angular velocity of wheel C

The angular velocity of wheel A (\(\omega_A\)) needs to be determined first. Use the given final angular velocity of wheel C (\(\omega_C\)). We are given:\[r_A = 10 \text{ cm} = 0.10 \text{ m}\]\[r_C = 25 \text{ cm} = 0.25 \text{ m}\]\[\omega_C = 100 \text{ rev/min}\]Convert \(\omega_C\) to radians per second:\[100 \text{ rev/min} \times \frac{2\pi \text{ radians/rev}}{60 \text{ s/min}} = \frac{200\pi}{60} \text{ radians/s} \approx 10.47 \text{ radians/s}\]Using the equality of tangential speeds:\[r_A \times \omega_A = r_C \times \omega_C\]\[0.10 \times \omega_A = 0.25 \times 10.47\]\[\omega_A = \frac{0.25 \times 10.47}{0.10} = 26.175 \text{ radians/s}\]
03

Determine the time required for wheel A to reach \(\omega_A\)

The rotational speed of wheel A increased from rest (initial angular velocity \(\omega_0 = 0\)) at a constant rate of \(1.6 \text{ rad/s}^2\). Using the formula for angular acceleration (\(\alpha\)):\[\omega = \omega_0 + \alpha t\]Substitute \(\omega_0 = 0\), \(\alpha = 1.6 \), and \(\omega = 26.175\):\[26.175 = 0 + 1.6 t\]Solving for time (\(t\)):\[t = \frac{26.175}{1.6} \approx 16.36 \text{ seconds}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity refers to how quickly an object spins around a central point. It's a measure of rotational speed and direction. In the exercise, wheel A has its angular velocity increased over time. The angular velocity is usually represented by the Greek letter \( \omega \). The formula for angular velocity is: \( \omega \ = \frac{\text{change in angle}}{\text{change in time}}\). Linear velocity, on the other hand, looks at speed along a straight path. When solving our exercise, we converted the final rotational speed of wheel C from revolutions per minute to radians per second. This conversion allows for more precise calculations.
Tangential Velocity
Tangential velocity is the speed of something moving along a circular path. Specifically, it鈥檚 the speed of a point on the edge of a wheel. For our problem, the belt ensures that the tangential velocities of wheels A and C are equal. The formula is: \( v \ = r \cdot \ \omega \), where \( r \) is the radius and \( \omega \) is angular velocity. We set the tangential velocities of both wheels equal to find the angular velocity of wheel A. Hence, by knowing the radii of both wheels and the final angular velocity of wheel C, we calculated wheel A鈥檚 angular velocity.
Angular Acceleration
Angular acceleration is how quickly the angular velocity changes. In our problem, it is given as \( 1.6 \ rad/s^2\). This means wheel A's rotational speed is ramped up at this constant rate. The formula to compute the time given angular acceleration is: \( \omega = \omega_0 + \alpha t \), where \( \omega_0 \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( t \) is the time. Since wheel A starts from rest, \( \omega_0 = 0 \). To find the time it takes for wheel A to reach its final angular velocity of 26.175 radians per second, we rearranged the formula and solved for \( t \).
Rotational Kinematics
Rotational kinematics deals with the motion of rotating objects without considering the forces causing the motion. It mirrors linear kinematics but applies to rotational movement. The common rotational kinematic equations include relationships between angular displacement, angular velocity, angular acceleration, and time. In our exercise, we used the equation \( \omega = \omega_0 + \alpha t \) from rotational kinematics to find how long it takes for wheel A to reach a certain angular velocity. Rotational kinematics also helped us determine the step-by-step relationship between the wheels' movements.

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Most popular questions from this chapter

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