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nchin
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A massless string is wrapped around a 10 kg cylinder. the picture will look like the letter b. you pull upward on the string with a force of P = 49N. The bottom of the cylinder s on horizontal table and will roll w/o slipping. using dynamics involving center of mass, find the magnitude and the direction of frictional force (f). (no slippage between string and cylinder)Solution:
Torque about cm: RP - Rf = Iα ---> R(P-f) = Iα
I = 1/2MR^(2)
α = a/R
R(P-f) = 1/2MR^(2) (a/R) ---> P-f = 1/2Ma
Forces Horz: f = ma
Why do we need to consider the forces horizontal?? Isn't it already included in the torque calculation?
P-ma = 1/2Ma
P = 3/2Ma
Ma = 2/3P
a = 2/3 (49N) / 10kg ---> 3.2667 N
My teacher got 32.667N? but how?
Torque about cm: RP - Rf = Iα ---> R(P-f) = Iα
I = 1/2MR^(2)
α = a/R
R(P-f) = 1/2MR^(2) (a/R) ---> P-f = 1/2Ma
Forces Horz: f = ma
Why do we need to consider the forces horizontal?? Isn't it already included in the torque calculation?
P-ma = 1/2Ma
P = 3/2Ma
Ma = 2/3P
a = 2/3 (49N) / 10kg ---> 3.2667 N
My teacher got 32.667N? but how?