Eduprobe
Question:
The minimum value of ω₀ below which the ring will drop down is
√3gμ(R−r)
√gμ(R−r)
√2g²μ(R−r)
√g²μ(R−r)
Solution:
N=Mω²(R−r)
f=Mg
f≤μN
Mg≤μMω²(R−r)
ω=√gμ(R−r)
√3gμ(R−r)
√gμ(R−r)
√2g²μ(R−r)
√g²μ(R−r)
N=Mω²(R−r)
f=Mg
f≤μN
Mg≤μMω²(R−r)
ω=√gμ(R−r)