Question 56
Time after which the cord breaks
Given:
m = 4.0 kg, l = 1.0 m, Tmax = 100 N, μk = 0.1.
m = 4.0 kg, l = 1.0 m, Tmax = 100 N, μk = 0.1.
Key idea: Kinetic friction acts tangentially and increases the speed of the disc. The cord tension acts radially and supplies the centripetal force.
1. Tangential acceleration
fk = μkmg
at = fk/m = μkg
Starting from zero tangential speed,
v = μkgt
2. Tension provides centripetal force
T = mv2/l
The cord breaks when T = Tmax.
Tmax = m(μkgt)2/l
3. Time of breaking
t = [1/(μkg)] × √Tmaxl/m
t = [1/(0.1 × 10)] × √(100 × 1)/4 = 5.0 s
t = 5.0 s
Note: The printed answer key appears to contain a dimensional typo. The correct general form is
t = [1/(μkg)] × √(Tmaxl/m).
The numerical answer is still 5.0 s because l = 1 m.