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# At t = 3.00 s a point on the rim of a 0.200-m-radius wheel ISBN: 9780321675460 31

## Solution for problem 28E Chapter 9

University Physics | 13th Edition

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Problem 28E

At t = 3.00 s a point on the rim of a 0.200-m-radius wheel has a tangential speed of 50.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.0 m/s2. (a) Calculate the wheel’s constant angular acceleration. (b) Calculate the angular velocities at t = 3.00 s and t = 0. (c) Through what angle did the wheel turn between t = 0 and t = 3.00 s? (d) At what time will the radial acceleration equal g?

Step-by-Step Solution:

Solution 28E Problem (a) Step 1: Radius of the rim r = 0.200 m Tangential speed v = 50.0 m/s Time t = 3.00 s Tangential acceleration a = -10.0m/s (slows down) Step 2: To calculate angular acceleration Relationship between angular and tangential acceleration a = r 10 = 0.2 = -50 rad/s 2 Problem (b) Step 1: To calculate angular velocities at t 3 = 3.0s and t 0 = 0 At t = 3s =3 r = 50 3 0.2 =3250 rad/s Step 2: At t = 0s, angular velocity = 0 We know that Angular acceleration = rate of change in angular velocity 3 0 = t3 0 Step 3: Rearranging the above equation = - (t t ) 0 3 3 0 =0250 - (-50)(3-0) 0 400 rad/s Problem (c) Step 1: To find the angular displacement made between t = 0 and t = 3.00 s ave= t t = 3 - 0 = 3 s = ave.t Step 2: 3 0 ave= 2 ave= 4002250 ave= 325 rad/s Step 3: Therefore = 325*3 = 975 rad Step 4: 1 rad = 1 revolution 2 = 975 2 = 155 revolution The wheel turned 155 rev in 3 seconds Problem (d) Step 1: To find the time when the radial acceleration becomes acceleration due to gravity(9.8 m/s ) v2 Radial acceleration a = r At a = g, v = g Step 2: v 2 g = g r v = gr g vg 9.8 0.2* vg 1.4 s

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##### ISBN: 9780321675460

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