A 0.70-kg disk with a rotational inertia given by MR 2/2 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 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 mass with a force of

The force will be "9.8 N".

Explanation:

The given values are:

mass,

m = 0.7 kg

M = 2

g = 9.8

Now,

⇒  

then,

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⇒  

⇒  

On substituting the values, we get

⇒      

⇒      

hence,

⇒  

On substituting the values, we get

⇒      

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⇒      

⇒      

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft