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A 0.5âkg block B slides without friction inside a slot cut in arm OA that rotates in a vertical plane. The rod has a constant angular acceleration Ëθ=10rad/s2. Knowing that when θ=45â and r=0.8m the velocity of the block is zero, determine at this instant, (a) the force exerted on the block by the arm, (b) the relative acceleration of the block with respect to the arm.

Respuesta :

a) Force exerted on the block by the arm is  4N.

b)  The relative acceleration of the block with respect to the arm is  10rad/s2.

(a) The force exerted on the block B by the arm OA is given by:

F = m × a = 0.5 kg × (10 rad/s2) × 0.8 m = 4 N

The force exerted on the block by the arm is equal to the product of the mass of the block and the relative acceleration of the block with respect to the arm. Since the velocity of the block is zero at the given instant, the relative acceleration of the block with respect to the arm is equal to the angular acceleration of the arm, i.e., 10rad/s2. Hence, the force exerted on the block by the arm is  

                               0.5kg×10rad/s2 =5N.

(b) The relative acceleration of the block with respect to the arm is given by:

a_rel = a + θ2 × r = 10 rad/s2 + (45 rad/s2)2 × 0.8 m = 15.6 m/s2

The relative acceleration of the block with respect to the arm is equal to the angular acceleration of the arm, i.e., 10rad/s2.

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