A block (mass = 2.0 kg) is hanging from a massless cord that is wrapped around a

Question

A block (mass = 2.0 kg) is hanging from a massless cord that is wrapped
around a pulley (moment of inertia = 1.1 × 10^-3kg · m2). Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.040 m during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

Explanation / Answer

(a) Newton’s 2nd law for the block gives the equation: SF = ma = mg - T Since the problem wants the angular acceleration and a = ra, then this becomes: mra = mg - T------------------->(1) The rotational analog of Newton’s 2nd law for the pulley is: St = Ia = rT From this we can solve for T: T = Ia/r--------------------------->(2) Plugging (2) into (1) and solving for a: a = mg / [mr +( I/r)] = (2kg)(9.8m/s²) / [(2kg)(0.04m) + (1.1 x 10?³kg•m² / 0.04m)] = 182.33rad/s (properly rounded) (b) The tension comes from either (1) or (2), I’ll use (1): mra = mg - T T = m(g - ra) = 2kg[9.8m/s² - (0.04m)(182.33rad/s²)] = 5N

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