A) Figure attached below
B) The linear speed of the coin = 0.59 m/s
C) Linear speed as coin begins to slip = 0.83 m/s
D) The tangential speed will remain the same as seen in part C
Given data :
mass of coin = 0.0050 kg
Distance of coin from the center of disk = 0.14 m
Time to make a complete revolution = 1.5 s
A) Diagram showing the vectors on the figure is attached below
B) Determine the Linear speed of the coin
Linear speed of coin = 2 * π * ( 0.14 ) / 1.5
= 0.59 m/s
C) Determine the linear speed of the coin when it just begins to slip
given that: friction between coin and disk = 0.50
Friction becomes maximum when coin begins to slip
Maximum frictional force (Fmax) = uV
where V = mg
- ∴ Fmax = u*mg ---- ( 1 )
- centripetal force = [tex]\frac{mv^{2} }{r}[/tex] ---- ( 2 )
Equating equations ( 1 ) and ( 2 ) to determine the linear speed ( v )
v² = u*r*g
∴ v = √(u*r*g ) = √( 0.5 * 0.14 * 9.8 )
= 0.83 m/s
D) If the experiment is repeated with a second coin glued to the top of the first coin the tangential speed will remain the same
Hence we can conclude that The linear speed of the coin = 0.59 m/s Linear speed as coin begins to slip = 0.83 m/s , The tangential speed will remain the same as seen in part C
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