A small coin of mass 80g is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration α = 2rad/s². The coefficient of static friction between the coin and the disc is µs = $\frac{3}{4}$ and coefficient of kinetic friction is µk = 0.5. The coin is placed at a distance r =1m from the centre of the disc. The magnitude of the resultant force on the coin exerted by the disc just before it starts slipping on the disc is:
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Question

A small coin of mass 80g is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration α = 2rad/s². The coefficient of static friction between the coin and the disc is µs = $\frac{3}{4}$ and coefficient of kinetic friction is µk = 0.5. The coin is placed at a distance r =1m from the centre of the disc. The magnitude of the resultant force on the coin exerted by the disc just before it starts slipping on the disc is:

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G1 Physics · Accepted Answer

Just before slipping, static friction is at its maximum: f=µₛmg=( 3 4 \frac{3}{4} 4 3 ​ )(0.08)(10)=0.6 N. The disc also exerts normal N=mg=0.8 N, so the resultant contact force is √(0.8²+0.6²)=1 N. Hence D.