A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is _____ N. (Take g = 10 ms–2)
The problem involves a system where a block of mass m=2 kg is connected to a uniform disc of mass M=4 kg and radius R=10 cm. The block is suspended and allowed to fall under gravity, causing the disc to rotate without slipping. We need to find the tension in the cord.
First, analyze the forces on the block.
Gravity acts downward: Fg=mg=2 kg×10 m/s2=20 N.
Tension T acts upward.
Applying Newton's second law to the block:
ma=mg−T (1)
Next, consider the rotational motion of the disc. The torque τ caused by the tension is given by:
τ=TR
The moment of inertia I of the disc is:
I=(1/2)MR2
Using the rotational form of Newton's second law:
τ=Iα
Since the linear acceleration a is related to the angular acceleration α by α=a/R, we have:
TR=(1/2)MR2(a/R)
Rearranging gives:
T=(1/2)Ma (2)
Substitute (2) into (1):
ma=mg−(1/2)Ma
Rearrange to solve for a:
a=(mg)/(m+(1/2)M)
Substitute m=2 kg, M=4 kg, and g=10 m/s2:
a=(2×10)/(2+(1/2)×4)=20/4=5 m/s2
Now find T using equation (2):
T=(1/2)×4×5=10 N
The computed tension is 10 N, fitting the expected range of (10,10).
