56 REASONING AND SOLUTION a If the wheel does not slip a point on the rim

# 56 reasoning and solution a if the wheel does not

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56. REASONING AND SOLUTION a. If the wheel does not slip, a point on the rim rotates about the axle with a speed v T = v = 15.0 m/s For a point on the rim ω = v T / r = (15.0 m/s)/(0.330 m) = 45.5 rad/s b. v T = r ω = (0.175 m)(45.5 rad/s) = 7.96 m/s 57. REASONING a. The constant angular acceleration α of the wheel is defined by Equation 8.4 as the change in the angular velocity, ω - ω 0 , divided by the elapsed time t , or ( ) 0 / t α ω ω = - . The time is known. Since the wheel rotates in the positive direction, its angular velocity is the same as its angular speed. However, the angular speed is related to the linear speed v of a wheel and its radius r by v = r ω (Equation 8.12). Thus, / v r ω = , and we can write for the angular acceleration that 0 0 0 v v v v r r t t r t ω ω α - - - = = = b. The angular displacement θ of each wheel can be obtained from Equation 8.7 of the equations of kinematics: 2 1 0 2 t t θ ω α = + , where ω 0 = v 0 / r and α can be obtained as discussed in part (a). SOLUTION a. The angular acceleration of each wheel is ( )( ) 2 0 2.1 m/s 6.6 m/s 1.4 rad/s 0.65 m 5.0 s v v r t α - - = = = - Chapter 8 Problems 423 b. The angular displacement of each wheel is ( ) ( ) ( ) 2 2 0 1 1 0 2 2 2 2 1 2 6.6 m/s 5.0 s 1.4 rad/s 5.0 s 33 rad 0.65 m v t t t t r θ ω α α   = + = +     = + - = + ______________________________________________________________________________ 58. REASONING For a wheel that rolls without slipping, the relationship between its linear speed v and its angular speed ω is given by Equation 8.12 as v = r ω , where r is the radius of a wheel. For a wheel that rolls without slipping, the relationship between the magnitude a of its linear acceleration and the magnitude α of the angular acceleration is given by Equation 8.13 as a = r α , where r is the radius of a wheel. The linear acceleration can be obtained using the equations of kinematics for linear motion, in particular, Equation 2.9. SOLUTION a. From Equation 8.12 we have that ( )( ) 0.320 m 288 rad /s 92.2 m/s v r ω = = = b. The magnitude of the angular acceleration is given by Equation 8.13 as α = a/r . The linear acceleration a is related to the initial and final linear speeds and the displacement x by Equation 2.9 from the equations of kinematics for linear motion; 2 2 0 2 v v a x - = . Thus, the magnitude of the angular acceleration is ( ) ( ) ( ) ( ) ( )( ) 2 2 0 2 2 2 2 2 0 / 2 92.2 m/s 0 m/s 34.6 rad /s 2 2 384 m 0.320 m v v x a r r v v x r α - = = - - = = = 59. REASONING AND SOLUTION a. If the rope is not slipping on the cylinder, then the tangential speed of the teeth on the larger gear (gear 1) is 2.50 m/s. The angular speed of gear 1 is then ω 1 = v / r 1 = (2.50 m/s)/(0.300 m) = 8.33 rad/s 424 ROTATIONAL KINEMATICS The direction of the larger gear is counterclockwise . b. The gears are in contact and do not slip. This requires that the teeth on both gears move with the same tangential speed. v T1 = v T2 or ω 1 r 1 = ω 2 r 2 So ( ) 1 2 1 2 0.300 m 8.33 rad/s 14.7 rad/s 0.170 m r r ω ω   = = =     The direction of the smaller gear is clockwise .  #### You've reached the end of your free preview.

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• Spring '16
• Ahmed
• Physics, Homework Key
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