Question

6/105 The semicircular disk of mass m= 2 kg 6/105 The semicircular disk of mass m = 2 kg is mounted in the light hoop of radius r = 150 mm and released from rest in position (a). Determine the angular velocity w of the hoop and the normal force N under the hoop as it passes position (b) after rotating through 180°. The hoop rolls without slipping. (a) (6) PROBLEM 6/105 ( Answer | N = mg (1+ 2)

6/105  The semicircular disk of mass m = 2 kg is mounted in the light hoop of radius r = 150 mm and released from rest in position (a). Determine the angular velocity ω of the hoop and the normal force N under the hoop as it passes position (b) after rotating through 180°. The hoop rolls without slipping.

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Answer 1

DATE to (oper) foc qr 30 my2 Io - Ic + more for me?! + m in mr2 = Ict mar T 37 2 my z 4 2 3 T » Ice m2 m/Ar) Ip= mx 16m2 + m ( r - Ax ) - XY 2 30 tmr² - 8mr? bid 3TT. t mod 2 3mr XY 37 3. Ip = mr( 976 TO) Now applying energy conservation, 2x Mxgx 4y = I xr? - 35-46) w2 = 99 = 1 x (916) W2 37 iste 6 PACE

169 x 65T FTN (95-16) - w? =) w2= 96 got X0.15 XC90-1) 969 3x8* GT-16) 32g 38 (977-16) w? - ما اے g 32 ř (9-16) Now, for normal reaction, assuming mass is concentrated cet C.G.Grid . N-mg - mrw?..... as C.G will be in circular motion. => N=mgtmxar xwe 3T =) Na mg + mx4 X 2 X 337236) - N= mgt mx 4 3T 32 15-16) N=mg+ 32 mg xq 327 (97-16) N = mgt 128 mg 35 (97-16) => IN= mg ( 1732 90T-16)) / classmate PAGE