Apply the rotational kinematic equations. Problem A wheel rotates with a constan

Question

Apply the rotational kinematic equations. Problem A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at t = 0, (a) through what angle does the wheel rotate between t = 0 and t = 2.00 s? Give your answer in radians and in revolutions. (b) What is the angular speed of the wheel at t = 2.00 s? (c) What angular displacement (in revolutions) results while the angular speed found in part (b) doubles? Strategy The angular acceleration is constant, so this problem just requires substituting given values into the proper equations. SOLUTION (a) Find the angular displacement after 2.00 s, in both radians and revolutions. Use the equation to the right, setting i = 2.00 rad/s, = 3.5 rad/s2, and t = 2.00 s. = it + 1 2 t2 = (2.00 rad/s)(2.00 s) + 1 2 (3.50 rad/s2)(2.00 s)2 = 11.0 rad Convert radians to revolutions. = (11.0 rad)(1.00 rev/2 rad) = 1.75 rev (b) What is the angular speed of the wheel at t = 2.00 s? Substitute the same values into the equation to the right. = i + t = 2.00 rad/s + (3.50 rad/s2)(2.00 s) = 9.00 rad/s (c) What angular displacement (in revolutions) results during the time in which the angular speed found in part (b) doubles? Apply the time-independent rotational kinematics equation. f2 i2 = 2 Substitute values, noting that f = 2i. (2 9.00 rad/s)2 (9.00 rad/s)2 = 2(3.50 rad/s2) Solve for the angular displacement and convert to revolutions. = (34.7 rad) 1 rev 2 rad = 5.52 rev LEARN MORE Remarks The result of part (b) could also be obtained using the equation 2 = i2 + 2 and the results of part (a). Question Suppose the radius of the wheel is doubled. Are the answers affected? If so, in what way? (Select all that apply.)

The angular speed at t = 2.00 s is smaller.

The angle rotated through from t = 0 to t = 2.00 s is smaller.

The angular speed at t = 2.00 s is greater.

The angular speed at t = 2.00 s is the same.

The angle rotated through from t = 0 to t = 2.00 s is greater.

The angle rotated through from t = 0 to t = 2.00 s is the same.

Explanation / Answer

(a) Just use the equations which are the rotational analog of the general kinematics equations:

= t + 0.5t²
= (2rad/s)(2.00s) + 0.5(3.50rad/s²)(2.00s)²
= 11rad

In revolutions:

11rad = 11rad(1.00rev/2rad)
= 1.751rev

(b) = + t
= 2rad/s + (3.50rad/s²)(2.00s)
= 9rad/s

(C) doubles   =18rad/s

2 - 02 =2

=((18^2)-(2^2))/(2*3.5)=45.71rad

45.71rad =45.71rad(1.00rev/2rad)
= 7.27rev

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