Answer:
9. (B) ¼ Mv²
10. (A) √(3gL)
11. 20 N
12. 5 m/s²
Step-by-step explanation:
9. The rotational kinetic energy is:
RE = ½ Iω²
RE = ½ (½ MR²) (v/R)²
RE = ¼ Mv²
10. Energy is conserved.
Initial potential energy = rotational energy
mgh = ½ Iω²
Mg(L/2) = ½ (⅓ ML²) ω²
g(L/2) = ½ (⅓ L²) ω²
gL = ⅓ L² ω²
g = ⅓ L ω²
ω² = 3g / L
ω = √(3g / L)
The velocity of the top end is:
v = ωL
v = √(3gL)
11. Sum of torques about the hinge:
∑τ = Iα
-(Mg) (L/2) + (T) (r) = 0
T = MgL / (2r)
T = (3.00 kg) (10 m/s²) (1.60 m) / (2 × 1.20 m)
T = 20 N
12. Sum of forces on the block in the -y direction:
∑F = ma
mg − T = ma
Sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a / R)
T = ½ Ma
Substitute:
mg − ½ Ma = ma
mg = (m + ½ M) a
a = mg / (m + ½ M)
Plug in values:
a = (3.0 kg) (10 m/s²) / (3.0 kg + ½ (6.0 kg))
a = 5 m/s²