4)Find the tension in left side of the cord if there is no slipping between the

Question

4)Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel.

5)Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.

4)Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel. 5)Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel. B if there is no slipping between the cord and the surface of the wheel. 3)Find the angular acceleration of the wheel A if there is no slipping between the cord and the surface of the wheel. 2)Find the linear accelerations of block {rm m}. Find the linear accelerations of block {rm{ {rm kg}}} cdot {rm{{rm m}}}^{rm{2}} , and the radius of the wheel be 0.110 {rm kg}, respectively, the moment of inertia of the wheel about its axis be 0.300 {rm kg} and 2.00 B be 4.00 A and uploaded image 1)Let the masses of blocks C if there is no slipping between the cord and the surface of the wheel.

Explanation / Answer

The mass of the block A is mA = 4.0 kg The mass of the block B is mB = 2.0kg the moment of inertia of the wheel I = 0.3 kg/m^3 The radius of the wheel r = 0.11 m ----------------------------------------------------------------- From Newton's laws             mA a = mA g - T1             T1 = mA g - mA a and for block B            mB a = T2 - mB g then            T2 = mB g + mB a now for pulley                        = I           (T1 - T2) R = I(a/R)             mAg - mA a -mBg -mB a = Ia /R^2            (mA - mB)g - (mA + mB)a = I a /R^2                       (mA - mB)g =  (mA + mB)a + I a /R^2           therefore the linear acceleration                a = (mA - mB) g / [(mA + mB) + (I/R^2) ]                            =  (4 -2) (9.8) / [(4 +2) + (0.3/(0.11)^2) ]                                   = 0.64 m/s^2 (2) The linear acceleration of the block B is also a = 0.64 m/s^2 Since the system as the same acceleration (3) The angular acceleration                = a/R = 0.64 / 0.11 = 5.82 rad/s^2 (4) The tension in the left side cord                                T2 = mB g + mB a                                       = (2kg) [ 9.8 + 0.64]                   = 20.88 N (5) The tension in the right side cord                T1 = mA g - mA a                       = (4) [ 9.8 - 0.64]                      = 36.64 N

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