IT IS NOT 15.77 ? * QO Safari File Edit View History Bookmarks Window Help , 52%

Question

IT IS NOT 15.77

? * QO Safari File Edit View History Bookmarks Window Help , 52% HET) Mon 11:11 PM 0 session.masteringphysics.com Ch. 12 HW Problem 12.70 8 of 10 > Constants Periodic Table ? Part A A3.0-kg disk of radius 47 mm rolls down a ramp inclined at an angle of 28° with the vertical. The disk starts out at rest and the coefficients of static and kinetic friction between the ramp and the disk are both 0.50 What is the rotational speed of the disk after it has traveled 1.5 m? -1 Submit Request Answer Provide Feedback Next>

Explanation / Answer

F_net = m a

m g sin28 - f = m a

and torque = I alpha

f r = (m r^2 /2)(a/r)

f = m a /2

putting in previous equation,

m g sin28 - ma/2 = m a

a = 2 g sin28 / 3 = 0.31 g

lets see if disk is rolling without slipping,

fs_max = us N = 0.50 m g cos28 = 0.44 m g

and f_required = m a / 2 = 0.15 m g

that is less than maximum value, so it will roll without slipping.

a = 0.31 x 9.8 = 3.04 m/s^2

vf^2 - vi^2 = 2 a d

v^2 - 0^2 = 2(3.04)(1.5)

v = 3.02 m/s

w = v/r = 3.02 / 0.047 m

w = 64.3 rad/s OR 10.2 s^-1

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