Solutions to Week 8 Assignment

1. Find the rotational inertia by using I = S_i=A^Bm_ir_i
   (a) I = m(r² + 0² + 0² + r²) = 2mr² = 2(3.0kg)(0.50m)² = 1.5kg·m²
   (b) I = m(0² + r² + 0² + r²) = 2mr² = 2(3.0kg)(0.50m/Ö2)² = 0.75kg·m²
   (c) I = m(r² + r² + r² + r²) = 4mr² = 4(3.0kg)(0.50m/Ö2)² = 1.5kg·m²
2. |t| = F_^r = mgr = (40.0kg)(9.8N/kg)(2.0m) = 780N·m
3. (a) w_f = w_i + aDt = 0 + aDt = aDt and t = Fr_^ = ma_tr_^= m(r_^a)r_^ = mr_^²a, so t = mr_^²w_f/Dt.
            t = [(182kg)(0.62m)²/30.0s](120rpm)(2prad/rev)(1min/60s) = 29N·m
   (b) a is constant, so w_av = (w_i + w_f)/2.
         W = tq = tw_avDt = t[(w_i + w_f)/2]Dt = t[(0 + w_f)/2]Dt = (29.3N·m)(120)(2p)930.0S)/2(60S) = 5.5kJ.
4. (a) Choose the axis of rotation at the point of contact between the vertical wall and the climber's feet.
          t_net = Tcosq(1.06m) - W_c(0.91m) = 0, so
          T = [(0.91m)W_c]/[(1.06m)cosq] = [(0.91m)(770N)]/[(1.06m)cos25°] = 730N.
   (b) SF_x = 0 = F_x - Tsinq
        SF_y = 0 = F_y + Tcosq  - W_c
    &nbssp;   So, F_x = Tsinq  and  F_y = W_c - Tcosq, and
        F = ÖF_x² + F_y²
           = ÖT²sin²q + W_c² + T²cos²q - 2W_cTcoosq
           = ÖT² + W_c² - 2W_cTcoosq
           = Ö(730N)² + (770N)² -2(770N)(730N)cos25°
     &nbbsp;     = 330N.
   Find the direction of F.
         q = tan^-1F_y/F_x = tan^-1(W_c - Tcosq )/(Tsinq) = tan^-1[(770N) - (730N)cos25°]/(730N)sin25° = 19°
   So, F = 330N at 19° above the horizontal.
5. (a) t_net = 0 = F_q(10.0cm)sin20.0° - F_w(41cm)sin30.0° - F_l(22cm)sin30.0°
   >      F_q = gsin30.0°[m_w(41cm) + m_L(22cm)]/(10.0cm)sin20.0° = (9.8N/kg)sin30.0°[(3.0kg)(41cm) + (5.0kg)(22cm)]/(10.0cm)sin20.0° = 330N.
   (b) t_net = 0 = F_q(10.0cm)sin20.0° - F_w(41cm)sin90.0° - F_l(22cm)sin90.0°
   >      F_q = g[m_w(41cm) + m_L(22cm)]/(10.0cm)sin20.0° = (9.8N/kg)[(3.0kg)(41cm) + (5.0kg)(22cm)]/(10.0cm)sin20.0° = 670N.
6. (a) I = 1/2MR² = 1/2(20.0kg)(0.224m)² = 0.502kg·m²
   (b) The torque required to overcome the friction must be added to that necessary to accelerate the wheel to 1200rpm in 4.00s in the absence of friction to get the total torque necessary to accelerate the wheel to 1200rpm in 4.00s.
        t = Ia + Ia_f = I(a + a_f) = 1/2MR²[Dw/Dt + (Dw/Dt)_f]
          = 1/2(20.0kg)(0.224m)²(1200rpm/4.00s + 1200rpm/60.0s)(2prad/rev)(1min/60s) = 17N·m.
7. (a) E_total = K_tr + K_rot = 1/2mv² + 1/2Iw² = 1/2mv² + 1/2(2/5mr²)(v/r)² =  1/2mv² + 1/5mv² = 7/10mv²
                  = 7/10(0.600kg)(5.00m/s)² = 10.5J.
   (b) Use energy conservation.
              DU = -DK
            mgh = K
                 h = K/mg = 10.5J/(0.600kg)(9.8N/kg) = 1.8m.

8. Use conservation of angular momentum
       L_i = I_iw_i = L_f = I_fw_f
       w_f = (I_i/I_f)w_i = 2.50/1.60(10.0rad/s) = 15.6rad/s.