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`        A sphere rolls down an inclined plane at 30°. Calculate its acceleration.`
2 years ago

```							On rolling down the slope the potential energy of the sphere decreases and kinetic energy increases also the total kinetic energy is equal to the sum of both rolling and translation kinetic energies.
So,
Loss of kinetic energy = gain of kinetic energy
Mgh = ½ MV2 + ½ I * (omega)2 [angular velocity]
now we know,  from the radius of gyration: I = M * K2
v = r (radius) * omega
Mgh = ½ MV2 + ½ (M*K2) * (v/r)2
2gh = v2 [ 1 + (K/r)2 ]
Now we are familiar with the equation
v2  – u2 = 2as, where u (initial velocity) =0
so, we get  a = v2 / 2s
where  ‘s’ is the total path (hypotanuse) which is at a height ‘h’
so    Sin (theta) = h / s
Sin (30 degrees) = h/s
½  = h/s
So  s = 2h
so from this we get :   a = v2 / 2(2h) = v2 / 4h
now substituting the value for v2
2gh = a(4h)  [ 1+ (K/r)2 ]
a = [2gh / 4h] [ 1 + (K/r)2  ]
so the equation for acceleration at   30° :a = [g / 2] [ 1 + (K/r)2  ]

```
2 years ago
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