- A block (mass = 2.0 kg ) is hanging from a mass less cord is wrapped around a

Question

- A block (mass = 2.0 kg ) is hanging from a mass less cord is wrapped around a pulley (moment of inertia =1.1 x10-3 kg 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 around the pulley remains constant at a value of 0.040 m during the block’s descent.
a)the acceleration of the falling block.
b) the angular acceleration of the pulley and
c ) 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

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.