- #1
MathewsMD
- 433
- 7
A 0.70-kg disk with a rotational inertia given by MR2/2 (M) is free to rotate on a fixed horizontal axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass (m) hangs from the free end. If the string does not slip, then as the mass falls and the cylinder rotates, the suspension holding the cylinder pulls up on the cylinder with a force of:
A. 6.9N
B. 9.8N
C. 16N
D. 26N
E. 29N
ans: B
My solution:
ma = mg - F
Fr = Ia/r
mg - ma = Ia/r2
mg - ma = Ma/2
a = (mg)/(m + 0.5M)
ƩFy = FN - Mg - Ma
FN = M (g + a) = ~9.77N
Please point out any errors since I really want to ensure I understood every process throughly and correctly. If you have anything to add, in terms of helpful steps or things to consider in general, please reply! :)
A. 6.9N
B. 9.8N
C. 16N
D. 26N
E. 29N
ans: B
My solution:
ma = mg - F
Fr = Ia/r
mg - ma = Ia/r2
mg - ma = Ma/2
a = (mg)/(m + 0.5M)
ƩFy = FN - Mg - Ma
FN = M (g + a) = ~9.77N
Please point out any errors since I really want to ensure I understood every process throughly and correctly. If you have anything to add, in terms of helpful steps or things to consider in general, please reply! :)